Question

A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.6 m/sm/s and at point B is 5.4 m/sm/s . Part A Use the impulse-momentum theorem to find how long the sled takes to travel from A to B.

Answer 1

Answer:

Time taken by the sled is 1.31 s

Solution:

As per the question:

Coefficient of kinetic friction, $\mu_{k} = 0.25$

Velocity at point A, $v_{A} = 8.6\ m/s$

Velocity at point A, $v_{B} = 5.4\ m/s$

Now,

To calculate the time taken by the sled to travel from A to B:

According to the impulse-momentum theorem, impulse and the change in the momentum of an object are equal:

Impulse, I = Change in momentum of the sled, $\Delta p$        (1)

$I = Ft$                                 (2)

where,

F = Force

t = time

p = momentum of the sled

Force on the sled is given by:

$F = \mu_{k}N$

where

N = normal reaction force = mg

where

m = mass of the sled

g = acceleration due to gravity

Thus

$F = \mu_{k}mg$                     (3)

Using eqn (1), (2) and (3):

$\mu_{k}mgt = m\Delta v$

$\mu_{k}gt = v_{A} - v_{B}$

$t = \frac{v_{A} - v_{B}}{\mu_{k}g}$

$t = \frac{8.6 - 5.4}{0.25\times 9.8}$

t = 1.31 s

Answer 2

Answer:

$\Delta t =1.31\ s$

Explanation:

given,

coefficient of kinetic friction, μ = 0.25

Speed of sled at point A = 8.6 m/s

Speed of sled at point B = 5.4 m/s

time taken to travel from point A to B.

we know,

J = F Δ t

J is the impulse

where  F is the frictional force.

t is the time.

we also know that impulse is equal to change in momentum.

$J = m(v_f - v_i)$

frictional force

F = μ N

where as N is the normal force

now,

$F\Delta t = m(v_f -v_i)$

$\mu m g \times \Delta t = m(v_f-v_i)$

$\mu g \times \Delta t = v_f-v_i$

$\Delta t =\dfrac{v_f-v_i}{\mu g}$

$\Delta t =\dfrac{8.6-5.4}{0.25\times 9.8}$

$\Delta t =1.31\ s$

time taken to move from A to B is equal to 1.31 s