Question

A horizontal force of 5.0 n pushes a 0.50-kg block against a vertical wall. the block is initially at rest. the frictional force is

Answer 1

There are 4 forces acting on the block:
- the horizontal force F=5.0 N that pushes the block against the wall
- the normal reaction N of the wall against the block, horizontal but in the opposite direction
- the friction force, equal to $F_f=\mu N$, pushing upwards
- the weight of the block: mg, pushing downward

If the block is in equilibrium, it means that F=N and that the friction force is equal to the weight of the block, so:
$F_f = mg=(0.50 kg)(9.81 m/s^2)=4.91 N$

Answer 2

The frictional force exerted on the block is 5 N.

Acceleration of the block

The acceleration of the block is calculated as follows;

$F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{0.5} \\\\a = 10 \ m/s^2$

Coefficient of kinetic friction

The coefficient of kinetic friction of the block is calculated as follows;

$\mu = \frac{a}{g} \\\\\mu = \frac{10}{9.8} \\\\\mu = 1.02$

Frictional force

The frictional force exerted on the block is calculated as follows;

$F_f = \mu F_n\\\\F_f = \mu mg\\\\F_f = \mu g m\\\\F_f = (a) m\\\\F_f = 10 \times 0.5\\\\F_f = 5 \ N$

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