Question

A 1.8 kg block is moved at constant speed over a surface for which coefficient of friction is equal to 1/4 . It is pulled by a force F acting at 45 degree with horizontal. The block is displaced by 2 m. Find the work done on the block by (a) the force F (b) friction (c) gravity?

Answer 1

m = 1.8 kg
μ = 0.25

F  acts at 45⁰ with horizontal.
s = displacement horizontally = 2 meters.
Frictional force along the horizontal in the direction opposite to the displacement
  = Ff = μ * N = μ m g = 0.25 * 1.8 kg * 10 m/s² = 4.5 Newtons
Work done by the Frictional force = Wf = Ff . s = 4.5 * (-2 meters) = - 9 Joules
   (it is negative as displacement is in the opposite direction to friction).

The component of force F, along the horizontal in the direction of s :
     F Sin 45⁰ = F/√2
Since, the block moves with a uniform speed, there is no net force on the block.  Hence
     F/√2 = Ff = 4.5 Newtons
     F = 4.5 √2  Newtons

Work done by F on the block = F . s = F/√2  * 2 = 9 Joules.
This is positive as the force does positive work on the block.

Gravitational force, ie., the weight and the normal force N are perpendicular to the displacement vector.  Hence, the dot product is 0, for the work done.

Answer 2

Answer:

Explanation:but the ans is 7.2jand -7.2j