Question

A box of canned goods slides down a ramp from street level into the basement of a grocery store with acceleration 0.84 m/s2 directed down the ramp. The ramp makes an angle of 42° with the horizontal. What is the coefficient of kinetic friction between the box and the ramp?

Answer 1

Answer:

0.785

Explanation:

Given:

Acceleration of the canned goods down the ramp (a) = 0.84 m/s²

Angle of inclination of ramp (x) = 42°

Acceleration due to gravity (g) = 9.8 m/s²

Let the coefficient of kinetic friction between the box and the ramp be 'μ'.

Draw a free body diagram of the box and label the forces along the ramp and perpendicular to the ramp.

From the diagram,

Normal force is given as:

$N=mg\cos x$

So, frictional force is given as:

$f=\mu N=\mu mg\cos x$

Also, from the diagram, the net force along the ramp is given as:

$F_{net}=mg\sin x-f=mg\sin x-\mu mg\cos x=mg(\sin x-\mu\cos x)$

Now, from Newton's second law, net force is equal to the product of mass and acceleration. So,

$F_{net}=ma\\\\mg(\sin x-\mu\cos x)=ma\\\\\sin x-\mu\cos x=\frac{a}{g}\\\\\mu\cos x=\sin x-\frac{a}{g}\\\\\mu\cos x=\frac{g\sin x-a}{g}\\\\\mu=\frac{g\sin x-a}{g\cos x}$

Now, plug in the given values and solve for 'μ'. This gives,

$\mu=\dfrac{(9.8)(\sin(42)-0.84}{(9.8)(\cos(42))}\\\\\\\mu=\dfrac{5.717}{7.283}\\\\\\\mu=0.785$

Therefore, the coefficient of kinetic friction between the box and the ramp is 0.785.