Question

A particle of mass2kg is released from rest and slides down a plane inclined at 30 degree to the horizontal . There is a constant resistance force of 4N. Find the speed of the particle after it has travelled 8 meter

Answer 1

Answer on Question#38258 – Physics – Mechanics

A particle of mass $2\mathrm{kg}$ is released from rest and slides down a plane inclined at 30 degree to the horizontal. There is a constant resistance force of $4\mathrm{N}$. Find the speed of the particle after it has travelled 8 meter

Solution:

First, we can write the Newton's second law along the plane $(F_{\text{resist}} = 4N)$:

$$F_{\text{net}} = \mathrm{ma}$$ $$F_{\text{net}} = (\mathrm{mg})_{\text{slope}} - F_{\text{resist}} = \mathrm{mg} \cdot \sin \alpha - F_{\text{resist}}$$

(2)in(1):

$$\mathrm{mg} \cdot \sin \alpha - F_{\text{resist}} = \mathrm{ma}$$

Acceleration of the car:

$$\begin{array}{l} a = \frac{\mathrm{mg} \cdot \sin \alpha - F_{\text{resist}}}{\mathrm{m}} = \mathrm{g} \cdot \sin \alpha - \frac{F_{\text{resist}}}{\mathrm{m}} = \\ = 9.81 \frac{\mathrm{m}}{\mathrm{s}^2} \cdot \sin 18.4{}^\circ - \frac{4\mathrm{N}}{2\mathrm{kg}} = 1.1 \frac{\mathrm{m}}{\mathrm{s}^2} \end{array}$$

Equation of motion for the car:

$$\begin{array}{l} \mathrm{d} = \frac{\mathrm{a t}^2}{2} \\ \mathrm{t} = \sqrt{\frac{2\mathrm{d}}{\mathrm{a}}} \end{array}$$

Rate equation for the car

$$V_1 = a t = a \cdot \sqrt{\frac{2\mathrm{d}}{\mathrm{a}}} = \sqrt{2\mathrm{d a}} = \sqrt{2 \cdot 8\mathrm{m} \cdot 1.1 \frac{\mathrm{m}}{\mathrm{s}^2}} = 4.2 \frac{\mathrm{m}}{\mathrm{s}}$$

Answer: the speed of the particle after it has travelled 8 meter is equal to $4.2\frac{\mathrm{m}}{\mathrm{s}}$.