Question

A 2kg block is initially at rest on a horizontal frictionless table. A force of 15N is then exerted on the block at an angle of 37° below the horizontal. The change in kinetic energy after moving 3m is?

Answer 1

Approximately
`36\; {\rm J}`.

Step-by-step explanation:

Under the assumption, there is no friction to hinder the motion of the block. As a result, the increase in the kinetic energy of the block would be equal to the external work done on the block.

Let
`F` denote this magnitude of this force, let
`s` denote the magnitude displacement, and let
`\theta` denote the angle between this force and displacement. To find the work that this force has done on the block, use the formula:

`\begin{aligned}(\text{work}) &= F\, s\, \cos(\theta)\end{aligned}`.

For example, since the block in this question is moving along a horizontal table, displacement of the block would be in the horizontal direction. It is given that the angle between this force and the horizontal direction is
`37^(\circ)`. Thus, the angle between the force and the displacement would be
`\theta = 37^(\circ)`.

Magnitude of this force is
`F = 15\; {\rm N}`, while magnitude of displacement is
`s = 3\; {\rm m}`. The work done on this block would be:

`\begin{aligned}(\text{work}) &= F\, s\, \cos(\theta) \\ &= (15\; {\rm N})\, (3\; {\rm m})\, \cos(37^(\circ)) \\ &\approx 36\; {\rm N\cdot m} = 36\; {\rm J}\end{aligned}`.

Hence, the kinetic energy of this block would increase by
`36\; {\rm J}`.