Question

A 11kg block slides up a 30° inclined plane at a constant velocity. The coefficient of friction between the block and the plane is μk=0.2. In order for the block to slide up the incline, someone must apply a force to the block. What is the magnitude of the applied force if it points horizontally?

Answer 1

Answer:

The magnitude of the applied force is 94.74 N

Explanation:

Mass of the block, m = 11 kg

Angle of inclination of the plane, $\theta = 30^{\circ}$

Friction coefficient, $\mu_{k} = 0.2$

Now,

Normal force that acts on the block is given by:

$F_{N} = mgcos\theta + Fsin\theta$           (1)

Now, to maintain the equilibrium parallel to ramp the forces must be balanced.

Thus

$Fcos\theta = \mu_{k}F_{N}$                       (2)

From eqn (1) and (2)

$Fcos\theta = \mu_{k}(mgcos\theta + Fsin\theta)$

$F(cos\theta - \mu_{k}sin\theta) = \mu_{k}mgcos\theta$

$F = \frac{\mu_{k}mgcos\theta}{cos\theta - \mu_{k}sin\theta}$

$F = \frac{0.2\times 11\times 9.8cos30^{\circ}}{cos30^{\circ} - 0.2\times sin30^{\circ}}$

F = 94.74 N