Susan's 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul's speed after being pulled 3.0 m.

Question

Accepted Answer

Hi, everyone in this practice problem. We're being asked to calculate a slat speed at the end of a run. We will have a sprinter pulling a slat off a mass of kg on a rough horizontal grass surface where the cord actually makes an angle of 20 degrees with the surface. The coefficient of kinetic friction between the slat and the grass is 0.2. And the athlete runs for a distance of 25 m pulling the court with a constant force of 100 and 20 Newton. We're being asked to calculate the slats speed at the end of the run by using the work energy principle. The options given are a 4.1 m per second. B 6.0 m per second, C 9.1 m per second and B m per second. Awesome. So the slat in this case will have to be modeled as a particle. And we are asked to find the speed by applying the work energy principle. So before we start doing that, I am going to start us off with drawing the free body diagram of our system. So we'll have the slot which is going to be represented by a sphere right here, which is essentially going to be a point like particle. The slot is going to be pulled by and athlete with a rope creating an angle of 20 degrees right here. The where the will equals to 20 degrees. I will draw the athlete right here pulling the sled and the slot itself will be experiencing a friction force with the rough surface of F K. So the slot is going to be pulled to the right. And therefore there will be a counter force F K which is the frictional force between the slat and the grass surface. And then from the court of the athlete pulling the slide, there will be a tension force which is pulling the slat to the right, which is represented by this right here. This the right here is going to be projected into the Y and the X axis. And I'm going to draw the projection to make it easier for us down the line to do our calculation. So the tension force is going to be projected along the Y axis or the vertical direction, which is then going to be t multiplied by sign of data because it is opposite to the known angle right here. While the horizontal one is going to be t multiplied by cosine of data just like. So no, now that we have analyzed the forces acting on our sphere or on our slat essentially on the horse of the direction on the vertical direction, there will be two different forces acting. The first one is going to be the weight of the slap itself, which is going to be W which essentially is just M multiplied by G. And along that, there will be a normal force acting which is acting upon as a result of the interaction between the slap and the grass surface, which is going to be represented by this end right here pointing upwards. So this will essentially be our free body diagram of our system. So the energy principle states that doing work on a system actually changes the system's energy. So when we have interaction via friction, the change in the system's energy is then going to actually be given by the following formula. So the total work done to the system W tote is going to equal to the total change in the kinetic energy delta K plus the total change in the thermal energy delta E T H. So delta K is the variation in the kinetic energy which is represented by half M V F squared minus half M V I squared just like. So N delta E T H is the variation of the thermal energy between the slap and the grass system and the W thought is going to be the network work done by the pension. So in this case, our delta E P H is actually going to equals to F K multiplied by the displacement delta X which is the friction force, F K multiplied by delta X. And our W total is actually going to be our tension force multiplied by delta X. Awesome. So now we will actually start by calculating our W total or the total work done by detention. The slap actually moves 25 m in the horizontal direction. So the total work done produced by the tension is going to be given by W total equals to C multiplied by delta X which is in the horizontal direction. The T is going to be T multiplied by cosine of theta multiplied that by the displacement which is delta X and that will give us a value of T equals to 120 Newton cosine theta is cosine of 20 degrees and the delta X or the displacement is known to be 25 m. All of this is known from the problem statement. So then W total is then going to equals to 19.1 Newton. Awesome. So now that we have found our total work done, we wanna next calculate our delta E P H. We want to note that the weight M G and the normal force N are perpendicular to the displacement. And thus the work is W total is not gonna be correlated to the weight nor the normal force. So the weight and the normal force is going to be coming in in our calculations for the delta E P H. So calculating the change in the thermal energy, the change in the thermal energy of the slap. And the graph system can be calculated by using F K multiplied by delta X F K is essentially mu K multiplied by the normal force multiplied by delta X here for delta E T H. And therefore, what we have to find first is our normal force because it is not given in our problem statement. So the way we want to find our normal force is by using Newton's second law in the right direction. Y there just like so and according to Newton's second law, Sigma F in a particular axis, which is in this case, in the Y axis will equals to in this particular uh problem is going to be equals to zero Newton because it is not moving at all or the slat is not moving at all in the vertical direction. And according to our free body diagram in the Y direction, there are three different forces acting upon our slat. The first one is our normal force. The next one is the projection of our tension. And the last one is our weight. So Sigma F Y is then going to equals to N minus M G plus T multiplied by sine of 20 degrees or sine of theta equals to zero. So then rearranging this to find the normal force, we can get normal force equals to M G minus T sine 20 degrees. And that will essentially be um M G D mass is going to be 45 kg. Known from the problem statement, the gravitational acceleration G is 9.81 m per second squared. And then the P is going to be 120 Newton multiplied that by sine of 20 degrees. So after the calculation, our normal force will then equals to 400.41 Newton just like so awesome. So now we can actually plug our normal force and then also our um friction, kinetic friction coefficient UK and our displacement to get delta E T H. So delta E T H is equals to mute K multiplied by normal force multiplied by delta X UK is given in the problem statement which is 0.2, the normal force is 400.41 Newton and the displacement displacement is 25 m and that will give us a delta E P H value of 2002.5 Newton just like some. So now in the final step of our calculation, what we wanna do is to plug everything back into the initial initial equation that we have here, which is W total equals delta K plus delta E T H. So I'm going to rewrite that equation right here. W total equals delta K plus delta E T H and opening everything up W total minus delta E T H will then equals to B delta K. The delta K is where we can find our slats final speed, which is what we are interested at looking. So W total minus delta E T H is then going to equals to 28 19. Newton minus 2002 point oh five. Newton equals to delta K, which is going to be half M V F squared minus half M V I squared. We know that the slot started from rest. So V I equals to zero. So the half M V I squared term will also equals to zero in this case. So therefore, our equation will then be 28 19.1 Newton minus 2002 point oh five Newton equals to half M V F squared. The left side is then going to equals two, 817 point oh five Newton. And then from there, we can actually rearrange things so that we have V F equals to the square root of two multiplied by 817 point oh five Newton divided by the mess. In this case, then R V F will equals to the square root of two multiplied by 817 point oh five Newton divided by the mass, which is given to be 45 kg. And that will essentially give us a final velocity value of six meters per second just like. So, so with the final speed of the slide being six m per second. That will actually be the answer to this particular practice problem, which will actually correspond to option B in our answer choices. So option B with the slot speed at the end of the run being six m per second is going to be the answer to this particular practice problem. And that will be all for this particular practice problem. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topic and that will be it for this one. Thank you.

Susan's 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul's speed after being pulled 3.0 m.

Verified Solution

Key Concepts

Work-Energy Principle

Friction and Normal Force

Components of Forces