Question

work and power problems

1. a student pushes a box with a force of 50 n across the floor for 6 m in the same direction as the force. how much work is done?
2. a worker lifts a 20 kg crate vertically upward by 2 m. how much work is done against gravity? (take g = 9.8 m/s²)
3. a mover applies a 120 n force at a 60° angle to the horizontal to push a crate 10 m along the floor. calculate the work done.
4. a motor does 5000 j of work in 25 s. what is the power output of the motor?

Explanation:

Step1: Recall work - formula for force in same direction as displacement

The formula for work when force $F$ and displacement $d$ are in the same direction is $W = F\times d$. Given $F = 50\ N$ and $d=6\ m$.
$W_1=F\times d=50\times6$

Step2: Calculate work for first case

$W_1 = 300\ J$

Step3: Recall work - formula against gravity

The force required to lift an object of mass $m$ against gravity is $F = mg$. Here, $m = 20\ kg$, $g=9.8\ m/s^{2}$ and $d = 2\ m$. First, find the force $F=mg=20\times9.8 = 196\ N$. Then, use the work - formula $W = F\times d$.
$W_2=F\times d=196\times2$

Step4: Calculate work for second case

$W_2 = 392\ J$

Step5: Recall work - formula for angled force

The formula for work when a force $F$ is applied at an angle $\theta$ to the displacement $d$ is $W=Fd\cos\theta$. Given $F = 120\ N$, $d = 10\ m$ and $\theta = 60^{\circ}$, and $\cos60^{\circ}=\frac{1}{2}$.
$W_3=Fd\cos\theta=120\times10\times\frac{1}{2}$

Step6: Calculate work for third case

$W_3 = 600\ J$

Step7: Recall power formula

The formula for power $P$ is $P=\frac{W}{t}$, where $W$ is work and $t$ is time. Given $W = 5000\ J$ and $t = 25\ s$.
$P=\frac{W}{t}=\frac{5000}{25}$

Step8: Calculate power for fourth case

$P = 200\ W$

Answer:

1. $300\ J$
2. $392\ J$
3. $600\ J$
4. $200\ W$