Question

a 12 kg box slides down a 30.0 degree ramp with an acceleration of 1.4 m/s^2. If the frictional force is 11N what is the applied force?

Answer 1

Given:

$$m = 12 \mathrm{~kg}$$ $$g = 9.8 \frac{m}{s^{2}}$$ $$\alpha = 30{}^{\circ}$$ $$F_{\text{friction}} = 11 \mathrm{~N}$$ $$a = 1.4 \frac{m}{s^{2}}$$ $$F_{\text{applied}} - ?$$

We have the equations of horizontal forces:

$$\begin{array}{l} m \cdot g \cdot \sin \alpha - F_{\text{friction}} - F_{\text{applied}} = m \cdot a \Rightarrow \\ \Rightarrow F_{\text{applied}} = m \cdot g \cdot \sin \alpha - F_{\text{friction}} - m \cdot a \\ \end{array}$$ $$F_{\text{applied}} = 12 \cdot 9.8 \cdot \sin 30{}^{\circ} - 11 - 12 \cdot 1.4 = 58.8 - 11 - 16.8 = 31 \mathrm{~N}.$$

So, the applied force is $31\mathrm{N}$ and its direction is the same as direction of the friction force (in the opposite direction of the motion).

Answer: 31 N.