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Box ramp

A 1 kg box started from rest downward an
1-meter high ramp inclined at 30 degrees. The
kinetic friction coefficient is 0.3. What was
its speed when it reaches the bottom, in m/s?

A. 3.1

B. 5.5

C. 4.8

D. 1.3

E. 4.4

1 Answer

3 votes

Answer:

(option A).

Step-by-step explanation:

The gravitational potential energy of the box at the top of the ramp is given by:

Ep = mgh

where m is the mass of the box, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ramp (1 m).

Ep = (1 kg)(9.8 m/s^2)(1 m) = 9.8 J

As the box slides down the ramp, its gravitational potential energy is converted into kinetic energy, which is given by:

Ek = (1/2)mv^2

where v is the velocity of the box at the bottom of the ramp.

The work done by the friction force on the box is given by:

W = f * d

where f is the force of friction and d is the distance traveled by the box along the ramp. The force of friction is given by:

f = μmg

where μ is the coefficient of kinetic friction (0.3) and mg is the weight of the box.

f = (0.3)(1 kg)(9.8 m/s^2) = 2.94 N

The distance traveled by the box along the ramp is:

d = h/sin(30°) = 2 m

Therefore, the work done by the friction force is:

W = (2.94 N)(2 m) = 5.88 J

The total mechanical energy of the box at the bottom of the ramp (neglecting air resistance) is equal to the sum of its kinetic energy and the work done by the friction force:

Ek + W = Ep

(1/2)mv^2 + 5.88 J = 9.8 J

Solving for v, we get:

v = sqrt[(2(9.8 J - 5.88 J))/m] = sqrt[(2(3.92 J))/1 kg] = 3.1 m/s

answered
User Marcos Cassaro
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