Question: A 20 kg box is sliding down a ramp of 30°. The frictional coefficient is k=0.1. What is (just) the frictional force on the box? Use the frictional force on a ramp formula. Remember: cos(30°) = 0.866 and sin(30°) = 0.5
A. 1 N
B. 16.97 N
C. 1.73 N
D. 9.8 N
1. **State the problem:** We need to find the frictional force acting on a 20 kg box sliding down a ramp inclined at 30° with a friction coefficient $k=0.1$.
2. **Formula for frictional force on a ramp:**
$$F_f = k \times N$$
where $N$ is the normal force.
3. **Calculate the normal force $N$:**
The normal force on an inclined plane is given by:
$$N = mg \cos(\theta)$$
where $m=20$, $g=9.8$, and $\theta=30^\circ$.
4. **Substitute values:**
$$N = 20 \times 9.8 \times 0.866 = 169.736$$
5. **Calculate frictional force:**
$$F_f = 0.1 \times 169.736 = 16.9736$$
6. **Round the answer:**
$$F_f \approx 16.97$$
7. **Interpretation:** The frictional force acting on the box is approximately $16.97$ N.
**Final answer:** B. 16.97 N