A 57 g tennis ball is dropped from rem Alter falling 18.0 m the ball has a speed
ID: 1290455 • Letter: A
Question
A 57 g tennis ball is dropped from rem Alter falling 18.0 m the ball has a speed of 12.0 m/s. Your tack is to determine the work done by the resistive forces during the ball's motion. a) What is included in the system that you chose? b) Describe what is happening in the initial and final states in terms of the object's motion and in terms of the other objects included in your system c) Describe the tota4lechanical energy of the system in the initial state mathematically. Then do the same for the final state d) How much work did the resistive forces do on the ball during the ball's motion? What is the meaning of the sign? 2. A spring with a 1.20 x 10^4 -N/m force constant is compressed 6.00 cm, and a 10.0-kg block is placed against it. When the spring is released, the block shoots forward along a horizontal surface whose coefficient of friction is 0.200. Follow the problem solving steps of 1 a) through c) to determine how far the block w ill navel before stopping. 3. A 900-kg car initially at rest rolls down a hill inclined at an angle of 5.00 degree. A 400-N friction force opposes its motion. Follow the problem solving steps of 1 a) through cl to determine the car's speed after moving 50 m down the hill.Explanation / Answer
Answer 1)
Weight of ball = 57g = 0.057 kg
Gravitational force = 0.57 N
Acceleration gravity = 10 m/sq.s
Let resistive force be F.
Net force = 0.57-F
Acceleration a = (0.57-F)/0.057
v2 = u2 + 2as
or 122 = 2*a*18
or a = 12*12/36 = 4 m/s2
0.57-F/0.057 = 4
or F = 0.342 N
Work done = 18 * 0.342 * cos(180o) = -6.156 J (ans)
Meaning of sign:
Negative sign means work done by resistive force is against motion or displacement.
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Answer 2)
When the system is released, mass will be in action of spring force till spring is free.
Then it shall decelerate.
Initial velocity u = 0 m/s
Compression = 6 cm
PE stored = 0.5*k*x2 = 0.5*1.20*104*0.062 = 21.6 J
PE is converted to (KE - Work done by friction)
Work done by friction (-Ff*x) = 0.2*10*10*0.06 = 1.2 J
KE = 21.6-1.2 = 20.4 J = 0.5*m*v2 or
v = 2.02 m/s
Final KE = 0
From time when spring force = 0, friction decelerates the body and brings to rest. Let it move 'X' m
So, Ff*X = 20.4 or X = 20.4/20 = 1.02 m
So, from release point, block moves 1.02+0.06 = 1.08 m before coming to halt (ans)
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Answer 3)
Angle of inclination = 5o
Force acting down the plane = mgsin(5o) = 784.4 N
Friction = 400 N
Net force = 384.4 N
Acceleration = 384.4/900 = 0.43 m/s2
After 50 metres moving,
v2 = u2 +2*0.43*50
or v = 6.56 m/s
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