Question

3. A 45 kg box is pushed up a 21 m ramp at a uniform speed. The top of the ramp is 3.0 m higher than the bottom. (i) What is the potential energy of the box at the top of the ramp relative to the bottom of the ramp? (ii) What work was done pushing the box up the ramp if friction were negligible? (iii) What work was done pushing the box up the ramp if the force of friction between the box and the ramp was 100. N?

Answer 1

1/2*m*v^2 = m*g*h
m = mass
v = speed
g = acceleration of gravity
h = height; more exactly, the
increase in height
Crossing mass out, it can be simplified to:
1/2*v^2 = g*h, then to
h = (v^2)/(2*g)
Let's say d is the distance. Then,
d = 2*h

Answer 2

Given that,

Mass of a box, m = 45 kg

Length of a ramp, l = 21 m

The top of the ramp is 3.0 m higher than the bottom.

Solution,

(a) The potential energy of the box at the top of the ramp relative to the bottom of the ramp is given by :

$P=mgh\\\\P=45\times 10\times 3\\\\P=1350\ J$

(b) The work done in pushing the box up the ramp is equal to the potential energy of the box at the top of the ramp relative to the bottom of the ramp i.e. 1350 J.

(c) The work done in pushing the  box up the ramp if the force of friction between the box and the ramp was 100 N is given by :

$W=mgh+f\times l$

f is frictional force

$W=1350+100\times 21\\\\W=3450\ J$

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Conservation of energy

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