Question

A block is sliding on a horizontal frictionless surface at a velocity of 12.0 m/s (E). A force of 25.0 N(S) acts on the block, causing it to accelerate to a velocity of 45.0 m/s (45 degree S of E). If the mass of the block is 7.50 kg, what is the work done by the force on the block?

Answer 1

Answer on Question #43395, Physics, Mechanics | Kinematics | Dynamics

A block is sliding on a horizontal frictionless surface at a velocity of $12.0 \, \text{m/s}$ (E). A force of $25.0 \, \text{N(S)}$ acts on the block, causing it to accelerate to a velocity of $45.0 \, \text{m/s}$ (45 degree S of E). If the mass of the block is $7.50 \, \text{kg}$ , what is the work done by the force on the block?

Solution:

In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force.

Work = Force * distance moved in direction of force

$$W = F \cdot \cos \theta \cdot d$$

In our case:

I give the formula for the Law of Cosines and use it to find the missing side length of a triangle.

$$c ^ {2} = a ^ {2} + b ^ {2} - 2 a b \cos \gamma$$

In our notations the final velocity of the block caused of the force action $\mathbf{v}_{\mathrm{F}}$ is:

$$v _ {F} ^ {2} = 4 5 ^ {2} + 1 2 ^ {2} - 2 \cdot 4 5 \cdot 1 2 \cdot \cos (4 5 {}^ {\circ}) = 1 4 0 5$$ $$v _ {F} = \sqrt {1 4 0 5} = 3 7. 4 8 \mathrm {m / s}$$

The force case causes to change the velocity from zero to $37.48\,\mathrm{m/s}$ on direction at angle $\theta$. To find angle $\theta$ we again use the Law of Cosines:

$$45^2 = 37.48^2 + 12^2 - 2 \cdot 37.48 \cdot 12 \cdot \cos(\theta + 90{}^\circ)$$ $$\begin{array}{l} \cos(\theta + 90{}^\circ) = \frac{37.48^2 + 12^2 - 45^2}{2 \cdot 37.48 \cdot 12} = -0.52945 \\ \theta + 90{}^\circ = \cos^{-1}(-0.52945) = 122{}^\circ \\ \theta = 32{}^\circ \\ \end{array}$$

The work is

$$W = F \cdot \cos \theta \cdot d$$

The kinematic equation that describes an object's motion is:

$$v_f^2 = 2a d$$

The symbol $d$ stands for the displacement of the object. The symbol $a$ stands for the acceleration of the object.

$$a = \frac{F \cdot \cos \theta}{m}$$ $$d = \frac{v_f^2}{2a} = \frac{v_f^2 m}{2F \cdot \cos \theta}$$

Thus,

$$W = \frac{m v_f^2}{2} = \frac{7.50 \cdot 37.48^2}{2} = 5267.8\,\mathrm{J}$$

Answer: $W = 5267.8\,\mathrm{J}$.

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