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SVEN [57.7K]
2 years ago
5

A 15.4 kg block is dragged over a rough, horizontal surface by a constant force of 182 N acting at an angle of 24◦ above the hor

izontal. The block is displaced 57.6 m, and the coefficient of kinetic friction is 0.135. The acceleration of gravity is 9.8 m/s 2.
Physics
1 answer:
Dima020 [189]2 years ago
3 0

Answer:

Explanation:

Considering Work done by friction is asked

Given

mass of block m=15.4\ kg

force F=182\ N

inclination \theta =24^{\circ}

block is displaced by s=57.6\ m

coefficient of kinetic friction \mu _k=0.135

Friction force F_r=\mu _kN

Normal reaction N=mg-F\sin \theta =15.4\times 9.8-182\sin (24)

N=76.9\ N

Friction Force F_r=0.135\times 76.9=10.38\ N

Work done by friction force

W=f_r\cdot s

W=10.38\times 57.6=597.92\ N                        

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Read 2 more answers
A car accelerates uniformly from rest to a speed of 6.6 m/s in 6.5 s. Find the acceleration the car presents during this time?
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Answer:

1.02 m/s²

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The following data were obtained from the question:

Initial velocity (u) = 0 m/s

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Acceleration can simply be defined as the change of velocity with time. Mathematically, it can be expressed as:

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Where:

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v is the final velocity.

u is the initial velocity.

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With the above formula, we can obtain the acceleration of the car as follow:

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Final velocity (v) = 6.6 m/s

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Therefore, the acceleration of the car is 1.02 m/s²

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