Question(II) How much work would movers do pushing a $46.0-\mathrm{kg}$ crate 10.3 m horizontally across a rough floor without acceleration, if the effective coefficient of friction was 0.40 ?

Studdy Solution

STEP 1

1. The crate has a mass of $46.0 \, \text{kg}$ .
2. The crate is pushed a distance of $10.3 \, \text{m}$ horizontally.
3. The coefficient of friction between the crate and the floor is $0.40$ .
4. The crate is moved without acceleration, implying constant velocity.
5. We are calculating the work done against friction.

STEP 2

1. Calculate the force of friction.
2. Calculate the work done against the friction force.

STEP 3

Calculate the force of friction.
The force of friction $f$ can be calculated using the formula: $f = \mu \cdot N$
where $\mu = 0.40$ is the coefficient of friction, and $N$ is the normal force. Since the crate is moving horizontally without acceleration, the normal force $N$ is equal to the gravitational force on the crate, which is $N = m \cdot g$ .
Given: - $m = 46.0 \, \text{kg}$ - $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity)
Calculate the normal force: $N = 46.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 450.8 \, \text{N}$
Now calculate the force of friction: $f = 0.40 \times 450.8 \, \text{N} = 180.32 \, \text{N}$

STEP 4

Calculate the work done against the friction force.
The work done $W$ is given by the formula: $W = f \cdot d$
where $d = 10.3 \, \text{m}$ is the distance the crate is pushed.
Substitute the known values: $W = 180.32 \, \text{N} \times 10.3 \, \text{m}$
Calculate the work done: $W = 1856.296 \, \text{J}$
The work done by the movers is approximately: $\boxed{1856.3 \, \text{J}}$

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