Part 2: Work Done to Speed Up an Object In this part of the lab, you will amount
ID: 2076369 • Letter: P
Question
Part 2: Work Done to Speed Up an Object In this part of the lab, you will amount of work done by a constant force the exerted to speed up the wireless force probe the horizontal direction. assume the e rolling friction are negligible. The constant force will be force provided by the force of tension exerted on the wireless force sensor. The tension will is caused the gravitational force exerted g mass. The velocity of the cart be measured by the motion sensor. The experimental set up is shown below 0.2 m Procedure: Pulling the wireless force sensor using a constant force 1. Make sure the track is horizontally leveled. Perform any fine adjustments by turning the leg screws on the track and check if the cart remains motionless when placed in the middle.Explanation / Answer
I am assuming you need help with the analysis questions.
Answer 1:
As you have written, the forces on the cart only are : The force due to tension of the string in +x direction, the force due to its own mass m in the downward direction,and the normal force of contact perpendicular to the horizontal surface.
If we consider the earth and the cart as a system, there do no exist any gravitational pull and normal force as external forces, as for this system, they become internal forces. The best example is to visualize yourself walking on the road. For an observer on earth, one will be the force along your acceleration direction and other two will be your weight and normal due to earth. However for an observer on moon, you are just on earth. All he can judge is the force along your acceleration. So similarly in this case, the force is along your direction of motion only ie. the tension force.
Answer 2:
The work energy theorem relates the net work done by the forces on a system to the change in kinetic energy.
For the cart as our system, we have forces along x direction as well as y direction. So our net work done is a sum of the 2 works Wx and Wy. Also Wx = mvx2 / 2 and Wy= mvy2 / 2.
So W = Wx + Wy = mvx2 / 2 + mvy2 / 2
This is the work energy statement for the cart as our object.
For cart and earth as our object, we have just one force along the x direction. so it is simply deducible that,
W = mvx2 / 2 . This is the work energy statement for this system.
Answer 3
For the cart as our system, we have 3 forces, 1 in x direction and 2 in y direction. Let the displacement along the x direction be 'x'. SO our work done due to the tensile force 'F' is Wx= Fx (as the force and displacement are in the same direction)
Now, let the final velocity of the system be vx. So KEx = mvx2 / 2
SO our work energy statement becomes Wx = KEx ie. Fx = mvx2 / 2 -(1)
This is the work done along x direction.
Similarly for normal and gravitational forces, the displacement is zero. So the work energy statement becomes,
0 = mvy2 / 2 which implies that vy = 0 ie. there is no motion along the y direction.
So the gravitational and normal forces do zero work.
The net work done on the cart is simply , W = Fx = 1.485 x 0.714 = 1.06 J
Answer 4:
Yes. This is because the tensile force remains unchanged in its magnitude or direction. So the net work done is the same. Also the other two forces have no contribution to the net work. So things remain unchanged.
Answer 5:
The KE of the cart can be calculated using the obtained velocity and the measured mass. Assuming the KE obtained is true ie. KE = 0.289 J.
Now clearly this is not equal to the work done on the cart. The reason for this discrepancy is in the fact that we assumed our surfaces to be frictionless. However, in reality, the surface and the wheels will have some factor of friction. So some amount of work done on the cart is used up in overcoming the friction during the motion. The friction acts along the negative x direction . Thus it changes the net force along the horizontal and correspondingly the net work done. Hence we get a difference in the two readings as we did not consider the frictional force and assumed it to be negligible.
Answer 6;
The Gravitational potential energy is given by PE = mgy .
Here in this case , net displacement in y direction is 0 . So change in PE = 0.
Also we derived earlier that Wy = 0. Which is in agreement with the above statement.
This solves all your problems. Kindly ask for any further doubts or clarifications. Happy to help.
Enjoy Physics !!