Question

A box slides down an inclined plane 37° above the horizontal. The mass of the block, m, is 35 kg, the coefficient of kinetic friction is 0.3 and the length of the ramp, d, is 8 m. (a) How much work is done by gravity? (b) How much work is done by the normal force? (c) How much work is done by friction? (d) What is the total work done?(1690J; 0; -671J; 1019J)

Answer 1

1690 J.

Step-by-step explanation:

The work done by gravity can be calculated using the formula:

W_gravity = mgh

where m is the mass of the box, g is the acceleration due to gravity, and h is the height of the inclined plane.

We can find the height h using trigonometry:

sin(37°) = h/d

h = sin(37°) * d

h = 0.6 * 8

h = 4.8 m

Now we can calculate the work done by gravity:

W_gravity = mgh

W_gravity = 35 kg * 9.81 m/s^2 * 4.8 m

W_gravity = 1690 J

Therefore, the work done by gravity is 1690 J.

Answer 2

(a) To calculate the work done by gravity, we can use the formula W = mgh (work equals the mass multiplied by gravity and the height). In this case, m is 35 kg, g is 9.81 m/s2, and h is the length of the ramp (8 m). Plugging this into the formula, W = 35(9.81)(8) = 2776.8 J.

(b) The work done by the normal force is 0 since it is perpendicular to the direction of motion.

(c) To calculate the work done by friction, we can use the formula W = Fs (work equals the friction force multiplied by the distance). In this case, F is the kinetic friction force (35*0.3 = 10.5 N) and s is the length of the ramp (8 m). Plugging this into the formula, W = -10.5(8) = -84 J.

(d) The total work done is the sum of the work done by gravity, the work done by the normal force, and the work done by friction. Thus, the total work done is 2776.8 + 0 + (-84) = 2692.8 J, which when rounded to the nearest integer is 2693 J.