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Force and Kinetic Energy

By 101jay Sep 30, 2012 814 Words
Part B
Now, suppose that Zak's younger cousin, Greta, sees him sliding and takes off her shoes so that she can slide as well (assume her socks have the same coefficient of kinetic friction as Zak's). Instead of getting a running start, she asks Zak to give her a push. So, Zak pushes her with a force of 125 \rm N over a distance of 1.00 \rm m. If her mass is 20.0 \rm kg, what distance d_2 does she slide after Zak's push ends? Remember that the frictional force acts on Greta during Zak's push and while she is sliding after the push. F= Fp-Fr

E= F*Lp= (Fp-Fr)*Lp= Fr*Lr
Lr= Lp*((Fp/Fr)-1)
Lr= 1*((125/(20*9.8*0.25))-1)= 1.6 m

Mark pushes his broken car 150 m down the block to his friend’s house. He has to exert a 110 N horizontal force to push the car at a constant speed. How much thermal energy is created in the tires and road during this short trip? thermal energy is created in the tires and road

= 110 * 150
=16500 J
A 30 kg child slides down a playground slide at a constant speed. The slide has a height of 4.0 m and is 7.0 m long. Using energy considerations, find the magnitude of the kinetic friction force acting on the child. The friction force F is parallel to the slope and is constant in magnitude, so its work is W = - F d

with d = length of the slide.

ΔU = m g Δh
- F d = m g Δh
F = - m g Δh / d = - 30 x 9.8 x (- 4.0) / 7.0 = 168N
A block of weight w = 15.0 N sits on a frictionless inclined plane, which makes an angle θ = 23.0° with respect to the horizontal, as shown in the figure. A force of magnitude F = 5.86 N, applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed.

Part A
The block moves up an incline with constant speed. What is the total work WTotal done on the block by all forces as the block moves a distance L = 3.40 m up the incline? Include only the work done after the block has started moving at constant speed, not the work needed to start the block moving from rest. Express your answer numerically in joules.

Since the block is moving at a constant speed there is no change in kinetic energy. Since work is the change in kinetic energy and there is no change, the net work on the block is zero. 0J

Part B

What is Wg, the work done on the block by the force of gravity w as the block moves a distance L = 3.40 m up the incline? Wg = -mgy = -15.0N*3.40m*sin(23.0) = -19.9J
-19.9 J

Top of Form

Weight (N): (input as the weight, which = mass * g)
Distance (m):
Angle (degrees):

Work (J):
Remember to use proper significant figures in your answer!
Bottom of Form

Part C

What is WF, the work done on the block by the applied force as the block moves a distance = 3.40m upthe incline? Since we found the net work to be zero (in Part A), the work done by the applied force has to offset the work done by gravity:

19.9 J

Part D
What is WN, the work done on the block by the normal force as the block moves a distance = 3.40 up the inclined plane? The normal force is perpendicular to the direction of motion, therefore it does no work on the block: 0 J

A 56 g ice cube can slide without friction up a 30 degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 10.0cm. The spring’s constant is 22 N/m. When the ice cube is released, what distance will it travel up the slope before reversing direction? When the ice cube is released, what distance will it travel up the slope before reversing direction? So, the kinetic energy possessed by the ice cube will be

=(1/2)*(k)*(x^2) where k is spring constant and x is the compression. =(1/2)*(22)*(0.1^2)
=0.11 J
Now as the cube is rising on the slope, there will come a time when the cube stops and all energy it has will be potential energy by virtue of gravity. And this potential energy will be equal to its initial kinetic energy. So, we have

mgh=initial kinetic energy=0.11 J
m=56g=0.056 kg; g=9.8 m/s^2
Solving the equation,
h=0.11/(0.056*9.8)=0.20043 m

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