Gravity Slope forces

Classical Mechanics Level 3

A block of 20 kg mass is pulled up a slope at constant velocity by applying force acting parallel to the slope. If the slope makes an angle of 30 degree with the horizontal, calculate

i) The work done in pulling the load up a distance of 3.0 meters along the slope.

ii) The increase in its potential energy of the block?

Details and Assumptions

Take $g = 9.8\text{ m/s}^2$
Provide the answer as (i) + (ii).

The answer is 588.

2 solutions

Rubayet Tusher
Apr 25, 2015

We have to consider the friction case to get the exact answer. We know, tan(theta)= "mu". So, we get (mu)=tan 30 = (1/sqrt3). Then, The total force required to neutralize the downward weight component & friction force becomes= mg(sin 30)+"mu"mg(cos 30) = (20)(9.8)=196 N. So, the net work done by this force is W= (196)(3) J=588 J. Well, Then, the height the block has achieved is (3)(sin 30) = 1.5. So, the increase in potential energy is = mg(1.5)=(20)(9.8)(1.5)=294 J. At last, We sum up these. The answer becomes = 588+294 =882 J

Friction is not mentioned, nor its coefficient. So we ignore it.

Kishore S. Shenoy - 5 years, 9 months ago

Vishnu Bhagyanath
Jul 8, 2015

Since nothing is mentioned about the roughness of the surface, we shall assume it is a smooth surface.

The force acting in the direction of the slope can be respresented as $mg \sin \theta$ .

$F = mg \sin\theta = 20 \times 9.8 \times \sin (30) = 98 N$ $W = F \times d = 98 \times 3 = 294 J$ Potential energy on the ground is $zero$ while potential energy at 3 meters will also be $294 J$ since all energy used to push it upto 3 meters is stored as potential energy.

$294 + 294 = 588 J$

Gravity Slope forces

The answer is 588.

2 solutions

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