Question

how much work did the movers do (horizontally) pushing a 41.0- kg crate 10.6 m across a rough floor without acceleration, if the effective coefficient of friction was 0.50

Answer 1

The work done by the movers pushing the crate horizontally is 2007.56 Joules.

Step-by-step explanation:

The work done by the movers pushing the crate horizontally can be calculated using the equation:

Work = Force x Distance

First, we need to find the force of friction acting on the crate:

Force of friction = Coefficient of friction x Normal force = 0.50 x (41.0 kg x 9.8 m/s^2)

Next, we can calculate the work done:

Work = Force of friction x Distance = (Force of friction) x (Distance)

Plugging in the values: Work = (0.50 x (41.0 kg x 9.8 m/s^2)) x 10.6 m

Simplifying: Work = 2007.56 Joules

Answer 2

The work done by the movers to push a 41.0 kg crate 10.6 m across a rough floor with a coefficient of friction of 0.50 is 2129.54 joules.

Step-by-step explanation:

To calculate the work done by the movers while pushing a 41.0 kg crate across a rough floor without acceleration, we will use the formula Work = force × distance.

Since the crate is not accelerating, the force applied by the movers is equal to the force of friction, which can be found using the formula force of friction = coefficient of friction × normal force. The normal force for a horizontal surface is equal to the weight of the crate, which can be found by multiplying its mass by the acceleration due to gravity (g = 9.8 m/s²).

• Therefore, the normal force (N) is 41.0 kg × 9.8 m/s² = 401.8 N.

• The force of friction (f) is 0.50 × 401.8 N = 200.9 N.

• The work done by the movers (W) is 200.9 N × 10.6 m = 2129.54 J (joules).

The work done by the movers to push the crate is 2129.54 joules.