Question 18: Laws of Motion

Question

A body of mass 10 kg is moving up on an inclined plane of $30^{\circ}$ with an acceleration $2 \mathrm{~ms}^{-2}$. Find the force required, if the coefficient of friction between the object and the plane surface is $\frac{\sqrt{3}}{6}$, [given $g=10 \mathrm{~ms}^{-2}$ ].

Accepted Answer

Forces along the plane
Ã¢ÂÂ DownÃ¢ÂÂslope component of weight:
$$m g\sin	heta = 10\cdot10\cdot\sin30^\circ = 100\cdot	frac12 = 50	ext{ N}$$
Ã¢ÂÂ Normal reaction:
$$N = m g\cos	heta = 10\cdot10\cdot\cos30^\circ = 100\cdot	frac{\sqrt3}{2} = 50\sqrt3	ext{ N}$$
Ã¢ÂÂ Friction (opposes upward motion):
$$f = \mu N = \frac{\sqrt3}{6}	imes50\sqrt3 = \frac{50\cdot3}{6} = 25	ext{ N}$$
NewtonÃ¢ÂÂs 2nd law up the plane
Let $P$ be the applied force up the plane. Then
$$P - \bigl(mg\sin	heta + f\bigr) \;=\; m a$$
$$P = m a + m g\sin	heta + f$$
Substitute numbers
$$P = 10\cdot2 \;+\;50 \;+\;25 = 20 + 50 + 25 = 95	ext{ N}$$
Answer: 95 N (Option D).

Answer

1.95 N

Answer

9.5 N

Answer

19 N

Answer

95 N

Laws of Motion

A body of mass 10 kg is moving up on an inclined plane of 30^{\circ} with an acceleration 2 \mathrm{~ms}^{-2}. Find the force required, if the coefficient of friction between the object and the plane surface is \frac{\sqrt{3}}{6}, [given g=10 \mathrm{~ms}^{-2} ].

Choose the correct answer: