Answer on Question #39827 – Physics – Mechanics
1. A 2400 W engine pulls a 200 kg block at constant speed up a 12.0 m long, 25.0° incline. Determine long does it takes to cover this distance.
where F ⃗ \vec{F} F is the pulling force. At the constant speed, the power of the engine is P = F ⋅ v P = F \cdot v P = F ⋅ v .
As the velocity of the block is constant, the acceleration equals to zero.
Then, write down this law in projectives:
{ m ⋅ 0 = m g sin α − F m ⋅ 0 = − m g cos α + N \left\{
\begin{array}{l}
m \cdot 0 = mg \sin \alpha - F \\
m \cdot 0 = -mg \cos \alpha + N
\end{array}
\right. { m ⋅ 0 = m g sin α − F m ⋅ 0 = − m g cos α + N
so the pulling force is F = m g sin α F = mg \sin \alpha F = m g sin α .
We can find the time of the motion of the block with the constant velocity:
t = l v = l : P F , t = ∣ m g sin α ∣ P . t = \frac{l}{v} = l: \frac{P}{F}, \quad t = \frac{ \left| mg \sin \alpha \right| }{P}. t = v l = l : F P , t = P ∣ m g sin α ∣ .
Let check the dimension.
[ t ] = m ⋅ k g ⋅ m s 2 W = m ⋅ N J / s = s . [t] = \frac{m \cdot kg \cdot \frac{m}{s^2}}{W} = \frac{m \cdot N}{J / s} = s. [ t ] = W m ⋅ k g ⋅ s 2 m = J / s m ⋅ N = s .
Let evaluate the quantity.
t = 12 ⋅ 200 ⋅ 9.8 ⋅ sin 25 ∘ 2400 = 4.14 ( s ) . t = \frac{12 \cdot 200 \cdot 9.8 \cdot \sin 25{}^\circ}{2400} = 4.14(s). t = 2400 12 ⋅ 200 ⋅ 9.8 ⋅ sin 25 ∘ = 4.14 ( s ) .
Answer: 4.14 s.