Question

An object has k.e. of 10J at a certain instant. If it is acted on by an opposing force of 5N, which of the number A to E below is the farthest distance it travels in metres before coming to rest?
A - 2
B - 5
C - 10
D - 20
E - 50

Answer 1

Answer:

d = 2 [m].

Explanation:

To solve this problem we must use the principle of conservation of energy, where the mechanical energy in a state plus the work done on the body, must be equal to the mechanical energy in the state E. This can be easily represented in the following equation.

$E_{A}+W_{A-E}=E_{E}$

where:

Ea = Mechanical energy in A [J]

Wa-e = Work among states a and e [J]

Ee = Mechanical energy in E [J].

The key to being able to understand this problem is that in state A, we only have kinetic energy, while the energy in state E is equal to zero (there is no movement). The work is equal to the product of force by distance, as work acts in the opposite direction to movement, this has a negative sign.

$10 - F*d = 0\\5*d = 10\\d = 10/5\\d = 2 [m]$