Answer to Question #212882 in Mechanics | Relativity for aiza

Question

The total energy required to move a 10-N block up a 25-degree incline for 10 meters is 500 Joule. A frictional force of 10-N opposes the motion of the block. Determine force (parallel the inclined) required to move the block, the rate of work done if the work was done in 10 seconds and the initial velocity of the block if its velocity at 10 meters is 30 m/s.

Answer 1

Explanations & Calculations

• If the needed parallel force is $\small F$, then the work done by that force should equal what is given as it is the only source doing work to energize the block.

$\qquad\qquad \begin{aligned} \small W=500J&=\small F\times10m\\ \small F&=\small \bold{50 N} \end{aligned}$

• That amount of work is done during a time period of 10s as the question says. Then the according to the definition of "rate of work": which is the power, it could be calculated.

$\qquad\qquad \begin{aligned} \small Power &=\small \frac{\text{work done}}{\text{time taken}}\\ &=\small \frac{500J}{10s}\\ &=\small \bold{50\,Js^{-1}\,or\,50\,W} \end{aligned}$

• This part could be done either by calculating the acceleration due to the net force & then applying any appropriate motion equation of the four or considering the energy conservation.
• Acceleration is easier, let's use energy conservation.
• Work done by the force F work against the friction & any other opposite force & change in the kinetic energy of the block.
• Then,

$\qquad\qquad \begin{aligned} \small 500J&=\small \Delta E_k+fs+mg\sin\theta.s\\ &=\small \frac{1}{2}.m(30^2-v_i^2)+10\times10+10.\sin25.10\\ \small 357.74&=\small 0.5\times \frac{10N}{9.8ms^{-2}}\times (30^2-v_i^2)\\ \small v_i&=\small \bold{14.1\,ms^{-1}} \end{aligned}$