By Work - Energy Theorem, Loss in K.E. = W.D. against the force × distance of friction or K.E. = μ mg S For constant K.E., S ∝ 1/m ∴ Truck will stop in a lesser distance.
By Work – Energy Theorem,
Loss in K.E. = W.D. against the force × distance of friction
If roads go straight up then angle of slope θ would be large so frictional forcef = μ mg cos θ would be less and the vehicles may slip. Also greater power would be required.
If roads go straight up then angle of slope θ would be large so frictional forcef = μ mg cos θ would be less and the vehicles may slip. Also greater power would be required.
Give an example in which a force does work on a body but fails to change its K.E.
When a body is pulled on a rough, horizontal surface with constant velocity. Work is done on the body but K.E. remains unchanged.
When a body is pulled on a rough, horizontal surface with constant velocity. Work is done on the body but K.E. remains unchanged.
See lessHow high must a body be lifted to gain an amount of P.E. equal to the K.E. it has when moving at speed 20 ms⁻¹. (The value of acceleration due to gravity at a place is 9.8 ms⁻²).
mgh = mv²/2 so h = 20.4 m
mgh = mv²/2
so h = 20.4 m
See lessIs it necessary that work done in the motion of a body over a closed loop is zero for every force in nature ? Why?
No. W.D. is zero only in case of a conservative force.
No. W.D. is zero only in case of a conservative force.
See lessA truck and a car moving with the same K.E. on a straight road. Their engines are simultaneously switched off which one will stop at a lesser distance?
By Work - Energy Theorem, Loss in K.E. = W.D. against the force × distance of friction or K.E. = μ mg S For constant K.E., S ∝ 1/m ∴ Truck will stop in a lesser distance.
By Work – Energy Theorem,
Loss in K.E. = W.D. against the force × distance of friction
or K.E. = μ mg S
For constant K.E., S ∝ 1/m
∴ Truck will stop in a lesser distance.
See lessMountain roads rarely go straight up the slope, but wind up gradually. Why?
If roads go straight up then angle of slope θ would be large so frictional forcef = μ mg cos θ would be less and the vehicles may slip. Also greater power would be required.
If roads go straight up then angle of slope θ would be large so frictional forcef = μ mg cos θ would be less and the vehicles may slip. Also greater power would be required.
See less