Work, Energy and Power
- Consider a car moving along a straight horizantal road with a speed of 72 km/h. If the coefficient of static friction between road and tyres is 0.5, the shortest distance in which the car can be stopped is
-
View Hint View Answer Discuss in Forum
Force due to friction = kinetic energy
μ mg s = 1 mv2 2
[ Here, v = 72 km/h= 72000 = 20 m/s ] 60 × 60 or , s = v2 = 20 × 20 = 40 m 2μg 2 × 0.5 × 10
Correct Option: B
Force due to friction = kinetic energy
μ mg s = 1 mv2 2
[ Here, v = 72 km/h= 72000 = 20 m/s ] 60 × 60 or , s = v2 = 20 × 20 = 40 m 2μg 2 × 0.5 × 10
- If the momentum of a body is increased by 50%, then the percentage increase in its kinetic energy is
-
View Hint View Answer Discuss in Forum
Initial momentum (p1) = p; Final momentum (p2) = 1.5 p and initial kinetic energy (K1) = K.
Kinetic energy (K) = p2 ∝ p2 2m or, K1 = p1 2 = p 2 = 1 K2 p2 1.5 p 2.25
or, K2 = 2.25 K.
Therefore, increase in kinetic energy is 2.25 K – K = 1.25 K or 125%.
Correct Option: C
Initial momentum (p1) = p; Final momentum (p2) = 1.5 p and initial kinetic energy (K1) = K.
Kinetic energy (K) = p2 ∝ p2 2m or, K1 = p1 2 = p 2 = 1 K2 p2 1.5 p 2.25
or, K2 = 2.25 K.
Therefore, increase in kinetic energy is 2.25 K – K = 1.25 K or 125%.
- The potential energy of a system increases if work is done
-
View Hint View Answer Discuss in Forum
When work is done upon a system by a conservative force then its potential energy increases.
Correct Option: D
When work is done upon a system by a conservative force then its potential energy increases.
- Two bodies of masses m and 4 m are moving with equal K.E. The ratio of their linear momenta is
-
View Hint View Answer Discuss in Forum
m1 = m, m2 = 4 m
K . E1 = K . E21 m1v12 = 1 m2v22 ; 1 mv12 = 1 mv22 2 2 2 2 v1 = 2 ⇒ v1 = 2v2 v2 Linear momentum of first body Linear momentum of second body = m1v1 = m .2v2 = 1 m2v2 4mv2 2
Correct Option: C
m1 = m, m2 = 4 m
K . E1 = K . E21 m1v12 = 1 m2v22 ; 1 mv12 = 1 mv22 2 2 2 2 v1 = 2 ⇒ v1 = 2v2 v2 Linear momentum of first body Linear momentum of second body = m1v1 = m .2v2 = 1 m2v2 4mv2 2
- A rubber ball is dropped from a height of 5m on a plane, where the acceleration due to gravity is not shown. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of
-
View Hint View Answer Discuss in Forum
According to principle of conservation of energy
Loss in potential energy = Gain in kinetic energy⇒ mgh = 1 mv2 ⇒ v = √2gh 2
If h1 and h2 are initial and final heights, then
v1 = √2gh1 , v2 = √2gh2
Loss in velocity , ∆v = v1 - v2 = √2gh1 - √2gh2
∴ Fractional loss in velocity= 1 - √0.36 = 1 - 0.6 = 0.4 = 2 5
Correct Option: B
According to principle of conservation of energy
Loss in potential energy = Gain in kinetic energy⇒ mgh = 1 mv2 ⇒ v = √2gh 2
If h1 and h2 are initial and final heights, then
v1 = √2gh1 , v2 = √2gh2
Loss in velocity , ∆v = v1 - v2 = √2gh1 - √2gh2
∴ Fractional loss in velocity= 1 - √0.36 = 1 - 0.6 = 0.4 = 2 5