Question

A car driver adjusts the pressure on car blraker so that the car travel at constant speed down a hill from P to Q the Magnitude of the chage in the car' s kinetic energy is ∆E the magnitude of the chage in the chage in its gravitation potential energy is ∆Ep Which statement is correct when the car reaches a.∆E,>E∆p. b∆E,=∆Ep. c∆E,>∆E,>0. d∆E,=0​

Answer 1

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When driver applies brakes and the car covers distance x before coming to rest, under the effect of retarding force F.

Then

2/1 mv^2

=Fx

or x= 2F/mmv^2

But when he takes turn, then

r/mm^2 = F

⇒r= F/mv^2

It is clear that x= 2r

i.e., by the same retarding force the car can be stopped in a less distance if the driver apply brakes. This retarding force is actually a friction force

Answer 2

Answer: When car reaches down the hill then ΔEp < ΔE

Step-by-step explanation:

Kinetic energy of a body is given by 1/2mv², where m is the mass of body and v is the speed.

When moving from P to Q, speed of car is kept constant. Hence, kinetic energy of car will remain constant.

∴ ΔE = 0

Potential energy of a body is given by mgh, where m is the mass of body, g is acceleration due to gravity and h is height at which object is placed.

When moving from P to Q, height of the car decreases. Hence, potential energy of car will decrease.

∴ ΔEp < 0

⇒ ΔEp < ΔE

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