A box (massm= 15kg) is placed on a slope (θ= 35◦), as shown in the diagram...

A box (massm= 15kg) is placed on a slope (θ= 35◦), as shown in the diagram below. Thecoefficients of friction between the box and the slope areμs= 0.6 andμk= 0.4. The box is released from reston the slope.(a) (2 points)Draw a free-body diagram for the box after it is released.(b) (8 points)Determine the acceleration of the box after it is released.

Expert Solution

a)

Mg - Weight of the box, M being the mass.
N - Normal force on the box
Ff - Frictional force
- Angle of the slope.

b)
The net force on the box,
M * a = M * g * sin - Ff ...(1)
Equating the forces perpendicular to the ramp, M * g * cos = N
Ff = k * N
Here we have taken the coefficient of kinetic friction, k since the gravitational force along the ramp is greater than the static frictional force.
Ff = k * M * g * cos ...(2)

Substituting (2) in (1),
M * a = M * g * sin - k * M * g * cos
Dividing M,
a = g * sin - k * g * cos
= g * [sin - k * cos]
= 9.81 * [sin(35) - 0.4 * cos(35)]
= 2.41 m/s^2

genius_generous answered 2 years ago

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