Download Q1: An object moves in a circle at constant speed. The work done by

Phys101
Term:131
Online HW-Ch07-Lec01
Q1:
An object moves in a circle at constant speed. The work done by the centripetal force
is zero because:
A.
B.
C.
D.
E.
Ans:
the displacement for each revolution is zero
the average force for each revolution is zero
there is no friction
the magnitude of the acceleration is zero
the centripetal force is perpendicular to the velocity
T
90ᵒ
v
�⃗ v and ∆x has the same direction
E ; W = ∆x�⃗. T
Q2:
At time t = 0 a 2.0 kg particle has a velocity of (4m/s) - (3m/s)j. At t = 3s its velocity
is (2m/s)i + (3m/s)j. During this time the work done on it was:
Ans:
v0 2 = 42 + (−3)2 = 16 + 9 = 25
v 2 = 22 + 32 = 4 + 9 = 13
1
1
1
1
W = ∆k = k − k 0 = mv 2 − mv0 2 = m(v 2 − v0 2 ) = × 2(13 − 25)
2
2
2
2
∴ W = −12 J
Q3:
A 0.50-kg object moves on a horizontal circular track with a radius
of 2.5 m. An external force of 3.0N, always tangent to the track,
causes the object to speed up as it goes around. If it starts from rest
its speed at the end of one revolution is:
Ans:
The work done by the force for small displacement dx is:
dw = dxf
Total work done W = �
2πr
0
Fdx = F �
2πr
0
dx = F 2πr
r
dx
F
∴ W = 2πrF = 2 × 3.14 × 2.5 × 3 = 13.7 m/s
KFUPM-Physics Department
1