Question

A mass of 500kg stands on a ramp inclined at 24degrees to the horizontal. The coefficient of friction
between the surfaces of the mass and ramp is 0.27. If the mass is allowed to move from rest, find the
distance moved by the mass in six seconds. Assume that g = 9.8 m/sec^2. Your answer should be correct to one decimal.​

Answer 1

Given :

Mass of object = 500 kg

Angle of inclination = 24°

Coefficient of friction = 0.27

Acc. due to gravity = 9.8 m/s²

To Find :

Distance covered by the object in six seconds.

Solution :

First of all we have to find acceleration of the object.

Net force acts on the object ::

$\sf:\implies\:F=mg\sin\theta-\mu N$

$\sf:\implies\:F=mg\sin\theta-\mu mg\cos\theta$

$\sf:\implies\:F=mg(\sin\theta-\mu\cos\theta)$

We know that, F = m a

$\sf:\implies\:ma=mg(\sin\theta-\mu\cos\theta)$

$\sf:\implies\:a=g(\sin\theta-\mu\cos\theta)$

• sin 24° = 0.4067
• cos 24° = 0.9135

$\sf:\implies\:a=9.8[0.4067-(0.27)(0.9135)]$

$\sf:\implies\:a=9.8[0.4067-0.2467]$

$\sf:\implies\:a=9.8\times 0.16$

$\bf:\implies\:a=1.568\:ms^{-2}$

Distance covered by object in the given interval of time can by calculated by using second equation of kinematics.

$\sf:\implies\:d=ut+\dfrac{at^2}{2}$

$\sf:\implies\:d=(0)(6)+\dfrac{(1.568)(36)}{2}$

$\sf:\implies\:d=\dfrac{56.448}{2}$

$:\implies\:\underline{\boxed{\bf{\orange{d=28.2\:m}}}}$

Answer 2

Answer:

Question :-

A mass of 500kg stands on a ramp inclined at 24degrees to the horizontal. The coefficient of friction

between the surfaces of the mass and ramp is 0.27. If the mass is allowed to move from rest, find the

distance moved by the mass in six seconds. Assume that g = 9.8 m/sec^2. Your answer should be correct to one decimal.

Solution :-

⟹d=ut+

2

at

2

\sf:\implies\:d=(0)(6)+\dfrac{(1.568)(36)}{2}:⟹d=(0)(6)+

2

(1.568)(36)

\sf:\implies\:d=\dfrac{56.448}{2}:⟹d=

2

56.448

:\implies\:\underline{\boxed{\bf{\orange{d=28.2\:m}}}}:⟹

d=28.2m

plz

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