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Chapter 4: Problem 419

How much is the work done in pulling up a block of wood weighing $2 \mathrm{KN}\( for a length of \)10 \mathrm{~m}$ on a smooth plane inclined at an angle of \(30^{\circ}\) with the horizontal? (A) \(1.732 \mathrm{KJ}\) (B) \(17.32 \mathrm{KJ}\) (C) \(10 \mathrm{KJ}\) (D) \(100 \mathrm{KJ}\)

Short Answer

Expert verified
The work done in pulling up a block of wood weighing \(2 \mathrm{KN}\) for a length of \(10 \mathrm{m}\) on a smooth plane inclined at an angle of \(30^{\circ}\) with the horizontal is \(10 \mathrm{KJ}\).

Step by step solution

01

Determine the component of weight along the inclined plane

We can find the force needed to pull the block up the inclined plane by calculating the component of the block's weight in the direction of the plane. The force required (F) can be found by the following formula: \[F = W \cdot sin(\theta)\] where W is the weight of the block and \(\theta\) is the angle of inclination. Step 2: Calculate the component of weight along the inclined plane
02

Calculate the component of weight

Given that the weight of the block (W) is \(2\mathrm{KN}\), and the angle of inclination \(\theta\) is \(30^{\circ}\), we can now plug these values into the formula from step one: \[F = 2\mathrm{KN} \cdot sin(30^{\circ})\] Since \(sin(30^{\circ}) = 0.5\), we have: \[F = 2\mathrm{KN} \cdot 0.5 = 1\mathrm{KN}\] Step 3: Calculate the work done in pulling up the block
03

Determine the work done using the force and distance

Now that we have found the force required to pull the block up the inclined plane, we can calculate the work done (W) using the following formula: \[W = F \cdot d\] where F is the force calculated in step 2, and d is the distance the block is pulled. Step 4: Calculate the work done
04

Calculate the work done

Given that the force required to pull the block is \(1\mathrm{KN}\) and the distance it is pulled is \(10\mathrm{m}\), we can now plug these values into the formula from step three: \[W = 1\mathrm{KN} \cdot 10\mathrm{m} = 10\mathrm{KJ}\] Based on our calculation, the work done in pulling up a block of wood weighing \(2 \mathrm{KN}\) for a length of \(10\mathrm{m}\) on a smooth plane inclined at an angle of \(30^{\circ}\) with the horizontal is \(10 \mathrm{KJ}\). Therefore, the correct answer is (C) \(10 \mathrm{KJ}\).

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Most popular questions from this chapter

A force \(\mathrm{F}=\mathrm{kx}\) (where \(\mathrm{k}\) is positive constant) is acting on a particle. Match column-I and column-II, regarding work done in displacing the particle. $$ \begin{array}{|l|l|} \hline \text { Column - i } & \text { Column - ii } \\ \hline \text { (a) From } \mathrm{x}=-4 \text { to } \mathrm{x}=-2 & \text { (P) Positive } \\ \hline \text { (b) From } \mathrm{x}=-2 \text { to } \mathrm{x}=-4 & \text { (Q) zero } \\ \hline \text { (c) From } \mathrm{x}=-2 \text { to } \mathrm{x}=+2 & \text { (R) negative } \\ \hline \end{array} $$ (A) \(\mathrm{a}-\mathrm{R}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{Q}\) (B) \(a-P, b-Q, c-R\) (C) \(a-R, b-Q, c-P\) (D) \(\mathrm{a}-\mathrm{Q}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{R}\)

The potential energy of a body is given by \(\mathrm{U}=\mathrm{A}-\mathrm{Bx}^{2}\) (where \(\mathrm{x}\) is displacement). The magnitude of force acting on the particle is (A) constant (B) proportional to \(\mathrm{x}\) (C) proportional to \(\mathrm{x}^{2}\) (D) Inversely proportional to \(\mathrm{x}\)

A metal ball of mass \(2 \mathrm{~kg}\) moving with a velocity of $36 \mathrm{~km} / \mathrm{h}$ has a head on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after the collision, the two balls move together, the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

A sphere collides with another sphere of identical mass. After collision, the two sphere move. The collision is inelastic. Then the angle between the directions of the two spheres is (A) Different from \(90^{\circ}\) (B) \(90^{\circ}\) (C) \(0^{\circ}\) (D) \(45^{\circ}\)

A body falls freely under the action of gravity from a height \(\mathrm{h}\) above the ground. $$ \begin{array}{|l|l|} \hline \text { Column - i } & \text { Column - ii } \\ \hline \text { (a) P.E. }=2 \text { (K.E.) } & \text { (P) constant at every point } \\ \hline \text { (b) P.E. }=\text { K.E. } & \text { (Q) at height }(\mathrm{h} / 3) \\ \hline \text { (c) P.E. }=(1 / 2) \text { (K.E.) } & \text { (R) at height }(2 \mathrm{~h} / 3) \\ \hline \text { (d) P.E.+ K.E. } & \text { (S) at height }(\mathrm{h} / 2) \\ \hline \end{array} $$ (A) \(a-P, b-Q, c-R, d-S\) (B) $\mathrm{a}-\mathrm{Q}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{S}, \mathrm{d}-\mathrm{R}$ (C) $\mathrm{a}-\mathrm{S}, \mathrm{b}-\mathrm{R}, \mathrm{c}-\mathrm{Q}, \mathrm{d}-\mathrm{P}$ (D) $\mathrm{a}-\mathrm{R}, \mathrm{b}-\mathrm{S}, \mathrm{c}-\mathrm{Q}, \mathrm{d}-\mathrm{P}$
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