A crate is sliding down a ramp that is inclined at an angle 42.4 ° above the horizontal. The coefficient of ki

Question

A crate is sliding down a ramp that is inclined at an angle 42.4 ° above the horizontal. The coefficient of kinetic friction between the crate and the ramp is 0.260. Find the acceleration of the moving crate.

Answer 1

Relevant forces are:

1. mg sin θ, the component of weight along the incline;
2. µN = µmg cos θ, force of friction, also along the incline, but in opposite direction to force above.

The component of weight normal to the surface is balanced by the incline surface itself.

Net force F = mg sin θ − µmg cos θ = mg(sin θ − µ cos θ).

Therefore, acceleration a

a = F/m = g(sin θ − µ cos θ).

Subst. numerical values,

a = 4.727 m/s².

Answer 2

F=ma

F= Fd - f
W=mg
Fd= W sin(42.4)
f= u N
f= u W cos (42.4)
F= mg(sin(42.4) - u cos(42.4))

a=F/m
a= g(sin(42.4) - u cos(42.4))
Just do the math

a= 9.81(sin(42.4) - 0.260 x cos(42.4))
a=4.7m/s
or
universally
a= 0.482g where g is acceleration due to gravity.