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raketka [301]
2 years ago
5

An object with a mass M = 250 g is at rest on a plane that makes an angle θ = 30 o above the horizontal. The coefficient of kine

tic friction between M and the plane is μk = 0.100. Mass M is attached by a string to another mass, m = 200 g, which hangs freely. When mass m has fallen 30.0 cm, its speed is:
(A) 83 cm/s
(B) 59 cm/s
(C) 160 cm/s
(D) 72 cm/s
(E) 48 cm/s
Physics
1 answer:
liubo4ka [24]2 years ago
3 0

Answer:

v = 79.2 m/s

Solution:

As per the question:

Mass of the object, m = 250 g = 0.250 kg

Angle, \theta = 30^{\circ}

Coefficient of kinetic friction, \mu_{k} = 0.100

Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

The tension in the string is given by:

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T        (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma

mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a

a = \frac{0.2\times 9.8 - 0.250\times 9.8\times sin30^{\circ} - 0.1\times 0.250\times 9.8\times cos30^{\circ}}{0.250 - 0.200}

a = 10.45\ m/s^{2}

Now, the speed is given by the third eqn of motion with initial velocity being zero:

v^{2} = u^{2} + 2ad

where

u = initial velocity = 0

Thus

v = \sqrt{2ad}

v = \sqrt{2\times 10.45\times 0.03} = 0.792\ m/s

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