Question

A 36.0 kg crate starting from rest is pulled across a floor with a constant horizontal force of 225 n. for the first 11.0 m the floor is frictionless, and for the nest 10.0 m the coefficient of friction is 0.20. what is the final speed of the crate after being pulled these 21.0 m

Answer 1

We can solve the problem by using conservation of energy.

In the first 11.0 meters, where the floor is frictionless, the kinetic energy gained by the block is equal to the work done by the horizontal force F:
$\Delta K_1 = Fd_1 = (225 N)(11.0 m)=2475 J$

In the second part of 10.0 m, the kinetic energy gained by the block is equal to the work done by the horizontal force F minus the work done by the frictional force:
$\Delta K_2 = F d_2 - \mu m g d_2=$
$=(225 N)(10.0m)-(0.20)(36.0 kg)(9.81 m/s^2)(10.0 m)=1544 J$

So, the total kinetic energy gained by the crate in these 21.0 meters is
$\Delta K=2475 J+1544 J=4019 J$
And since the crate started from rest, this is equal to the final kinetic energy of the crate, so we can calculate its final velocity starting from this value:
$K_f = \frac{1}{2} m v_f^2$
$v_f = \sqrt{ \frac{2K_f}{m} } = \sqrt{ \frac{2 \cdot 4019 J}{36 kg} }=14.9 m/s$

Answer 2

Final speed of the crate after being pulled these 21 m is 14.9 m/sec.

Given :

Mass = 36 kg

Horizontal Force = 225 N

For the first 11 m the floor is frictionless and for the next 10 m the coefficient of friction is 0.20.

Solution :

We know that the change in kinetic energy is,

$\rm \Delta K=\dfrac{1}{2}mv^2$  ---- (1)

Now,

$\rm \Delta K_1 = F\times D_1 =225\times11=2475\;J$

$\rm \Delta K_2 = F\times D_2 - \mu mgD_2$

$\rm \Delta K_2 = 225\times 10- 0.20\times 36 \times 9.81 \times 10$

$\rm \Delta K_2 = 1544\; J$

Therefore,

$\rm \Delta K = \Delta K_1 + \Delta K_2$

$\rm \Delta K = 2475 + 1544$

$\rm \Delta K = 4019\;J$

Now from equation (1),

$\rm 4019 = \dfrac{1}{2}\times36\times v^2$

$\rm v = \sqrt{223.278}$

$\rm v =14.9\;m/sec$

Therefore, final speed of the crate after being pulled these 21 m is 14.9 m/sec.

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