A 0.171 kg screwdriver initially at rest,
slides down a ramp that is inclined
at .
What is the magnitude of the normal
force of the screwdriver?

Ask by Chandler Valdez. in the United States

Mar 22,2025

Question

A 0.171 kg screwdriver initially at rest, slides down a ramp that is inclined at $$ 57.5^{\circ} $$. What is the magnitude of the normal force of the screwdriver? $$ \mathrm{n}=[\text { ? ] N } $$

UpStudy AI · Accepted Answer

The gravitational force acting on the screwdriver is given by

$$
\text{Weight} = m g
$$

where 
$$
m = 0.171 \, \text{kg} \quad \text{and} \quad g = 9.8 \, \text{m/s}^2.
$$

Since the normal force on an inclined plane is the component of the weight perpendicular to the surface, we have

$$
n = m g \cos(\theta)
$$

with the angle of inclination

$$
\theta = 57.5^\circ.
$$

Step 1. Calculate $$m g$$:

$$
m g = 0.171 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1.6758 \, \text{N}.
$$

Step 2. Compute $$\cos(57.5^\circ)$$:

Using a calculator,

$$
\cos(57.5^\circ) \approx 0.536.
$$

Step 3. Determine the normal force:

$$
n = 1.6758 \, \text{N} \times 0.536 \approx 0.8996 \, \text{N}.
$$

Rounded to two significant figures, the magnitude of the normal force is

$$
n \approx 0.90 \, \text{N}.
$$
The normal force is approximately 0.90 N.

A 0.171 kg screwdriver initially at rest,
slides down a ramp that is inclined
at .
What is the magnitude of the normal
force of the screwdriver?

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A 0.171 kg screwdriver initially at rest, slides down a ramp that is inclined at . What is the magnitude of the normal force of the screwdriver?