Coal is lifted out of a mine a vertical distance of 50 m by an engine that supplies 500 W to a conveyer belt. How much coal per minute can be brought to the surface? Ignore the effects of friction.

61 kg

Explanation:

Given data

distance (d) : 50 m

power (P) : 500 W = 500 J/s

mass (m) : ?

time (t) : 1 min = 60 s

The engine must do some work (w) to lift the coal. The work done in 1 minute is:

P = w / t

w = P. t = 500 J/s * 60 s = 3.0 * 10⁴ J

The work is equal to the force exerted (F) times the distance (d).

w = F * d

F = w / d = 3.0 * 10⁴ J / 50 m = 6.0 * 10² N

The force exerted is equal to the mass lifted (m) times the acceleration. Here, the acceleration to surpass is that of gravity (g = 9.8 m/s²).

F = m * g

m = F / g = 6.0 * 10² N / 9.8 m/s² = 61 kg

mass = 61.16 kg

Explanation:

given data

distance = 50 m

energy = 500 W = 500 * 60sec = 30000 J/min

to find out

mass

solution

we will apply here energy equation

that is

energy = m*g*h ... 1

put here all value m mass here and g = 9.81 and h is distance

energy = m*g*h

30000 = m*9.81*50

mass = 61.16 kg