A 50 kg wooden crate is slid across a flat surface for 60 meters by a force of 90.0 N. If the coefficient of sliding friction is

Question

4 years ago

6

9.15, what will be the final speed of the crate?

1 answer:

Alika [10] · Answer 1 · 2021-12-04T08:08:32+03:00

The final speed of the crate is 6.3 m/s

Explanation:

Please note that there is a typo in the text of the question: the coefficient of sliding friction is 0.15, not 9.15.

First of all, we need to find the force of friction acting on the crate. This is given by:

$F_f = \mu mg$

where:

$\mu=0.15$ is the coefficient of friction

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F_f = (0.15)(50)(9.8)=73.5 N$

Now we can find the acceleration of the crate by using Newton's second law:

$F-F_f = ma$

where

F = 90.0 N is the force applied forward on the crate

$F_f = 73.5 N$ is the force of friction, acting backward

a is the acceleration of the crate

Solving for a,

$a=\frac{F-F_f}{m}=\frac{90.0-73.5}{50}=0.33 m/s^2$ in the forward direction.

Finally, we can find the final speed of the crate by using suvat equations:

$v^2-u^2 = 2as$

where:

v is the final speed

u = 0 is the initial speed

$a=0.33 m/s^2$ is the acceleration

s = 60 m is the distance covered

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(0.33)(60)}=6.3 m/s$

Learn more about friction, forces and Newton's second law here:

#LearnwithBrainly