Question

The coefficient of kinetic friction between a 30-kg box and a tile floor is 0.31. What is the applied force if the box is accelerating at 5 m/s/s?

Answer 1

• Applied force is 243N.

Step-by-step explanation:

Given that a mass of 30kg is accelerating at 5m/s² on a surface which has coefficient of kinetic friction = 0.31.

• The forces acting of the box will be applied force by an agent, normal reaction, gravitational force, and frictional force.
• Normal reaction and weight of the body will cancel each other.
• Frictional force is numerically given as, μN, where N is the Normal reaction.
• In this case N = mg .
• Let the applied force is F .
• Frictional force will oppose the relative motion between the block and the surface, so it will act in a direction opposite to the applied force.

So we can write the equation as,

→ Net force = mass * acceleration

→ F - μN = 30kg * 5m/s²

→ F - 0.31 * 30kg * 10m/s² = 150kg-m/s² ( g≈10m/s²)

→ F - 93N = 150N

→ F = 150N + 93N

→ F = 243N

Hence the applied force is 243N .