Question

A 75 kg box slides down a 25.0 degree ramp with an acceleration of 3.60 m/s^2.

Answer 1

Answer:

The coefficient and acceleration are 0.061 and 8.284 m/s².

Explanation:

Given that,

Mass = 75 kg

Angle = 25.0°

Acceleration = 3.60 m/s²

We need to find the force

$F=ma$

$F=75\times3.60$

$F=270\ N$

Now, we need to calculate the parallel force

$F'=mg\sin\theta$

$F'=75\times9.8\times\sin25^{\circ}$

$F'=310.62\ N$

We calculate the friction force

$f_{\mu}=F'-F$

$f_{\mu}=310.6-270$

$f_{mu}=40.6\ N=approx 41\ N$

If the box is moving in an angle, then the weight

$W = mg\ \cos\theta$

If it is horizontally then

$W = mg$

So, The normal force is

$F'' = 75\times9.8\times\cos25^{\circ}$

$F''= 666.13\ N$

We need to calculate the coefficient between the box and ramp

$f_{\mu}=\mu\times F''$

Where, $\mu$=friction coefficient

$f_{\mu}$=frictional force

F''=normal force

Put the value into the formula

$\mu=\dfrac{f_{\mu}}{F''}$

$\mu=\dfrac{41}{666.13}$

$\mu=0.061$

We need to calculate the acceleration

$F''-f_{\mu}=ma$

$mg\cos25^{\circ}-\mu mg=ma$

$175\times9.8\cos25.0^{\circ}-0.061\times175\times9.8=175\times a$

$a=\dfrac{175\times9.8\cos25.0^{\circ}-0.061\times175\times9.8}{175}$

$a=8.284\ m/s^2$

Hence, The coefficient and acceleration are 0.061 and 8.284 m/s².

Answer 2

Below is the solution:

angle = 25 deg 
m = 75kg 
a = 3.6m/s/s 
F[x] = 75(9.8) sin(25) = 310.62N 
f[y] = 75(9.8) cos(25) = 666.1N 
f=ma 
310.62 - 666.1u = 75(3.6) 
666.1u = 310.62 -270= 40.62 
u = 0.061 

175(9.8) sin(25) - 175(9.8)(0.061)cos(25) = 175a 
175a = 724.79 -95.587= 629.203 
a = 3.595 m/s/s