WS1: Work Done by a Constant Force & Dot Product

Work Done by a Constant Force & Dot Product Worksheet

This is a worksheet based on lecture “Work Done by Constant Force and Dot Product“

Level 1 Practice Worksheet

Questions

Q1. The SI unit of work done is equivalent to:

(a) $\text{kg}\cdot\text{m/s}$

(b) $\text{kg}\cdot\text{m}^2/\text{s}^2$

(c) $\text{kg}\cdot\text{m}/\text{s}^2$

(d) $\text{kg}\cdot\text{m}^2/\text{s}^3$

Q2. The dimensional formula for work done is:

(a) $[\text{M L T}^{-1}]$

(b) $[\text{M L}^2 \text{T}^{-2}]$

(c) $[\text{M L}^{-1} \text{T}^{-2}]$

(d) $[\text{M L}^2 \text{T}^{-1}]$

Q3. Work done is a scalar quantity because it is defined as:

(a) The vector sum of force and displacement

(b) The ratio of force to displacement

(c) The cross product of force and displacement

(d) The dot product of force and displacement

Q4. A body is displaced by $5\text{ m}$ under the action of a constant force of $10\text{ N}$ acting in the direction of displacement. The work done by the force is:

(a) $2\text{ J}$

(b) $15\text{ J}$

(c) $50\text{ J}$

(d) $0\text{ J}$

Q5. A force of $20\text{ N}$ acts on a particle, displacing it by $4\text{ m}$ at an angle of $60^\circ$ to the direction of the force. The work done is:

(a) $80\text{ J}$

(b) $40\text{ J}$

(c) $20\text{ J}$

(d) $0\text{ J}$

Q6. A porter carries a heavy suitcase on his head and walks horizontally on a level platform at a constant speed. The work done by the force of gravity on the suitcase is:

(a) Positive

(b) Negative

(c) Zero

(d) Depends on the weight of the suitcase

Q7. Work done by kinetic friction is generally:

(a) Always positive

(b) Always zero

(c) Negative

(d) Infinite

Q8. A force $\vec{F} = (2\hat{i} + 3\hat{j})\text{ N}$ produces a displacement $\vec{s} = (4\hat{i} – 2\hat{j})\text{ m}$. The work done is:

(a) $2\text{ J}$

(b) $14\text{ J}$

(c) $8\text{ J}$

(d) $-6\text{ J}$

Q9. A particle moves from position $\vec{r}_1 = (1\hat{i} + 2\hat{j})\text{ m}$ to $\vec{r}_2 = (4\hat{i} + 6\hat{j})\text{ m}$ under a force $\vec{F} = (3\hat{i} + 2\hat{j})\text{ N}$. The work done by the force is:

(a) $17\text{ J}$

(b) $26\text{ J}$

(c) $8\text{ J}$

(d) $12\text{ J}$

Q10. If two vectors $\vec{F} = (3\hat{i} + c\hat{j})\text{ N}$ and $\vec{s} = (2\hat{i} – 4\hat{j})\text{ m}$ result in zero work done, the value of $c$ is:

(a) $1.5$

(b) $-1.5$

(c) $2.0$

(d) $-2.0$

Q11. Work done by a centripetal force on a particle moving in a uniform circular path is always zero because:

(a) Displacement is zero in one full revolution only

(b) Force is perpendicular to velocity at every instant

(c) Force acts along the tangent

(d) Magnitude of force is zero

Q12. If the angle between the force vector and displacement vector is $120^\circ$, the work done is:

(a) Positive

(b) Zero

(c) Negative

(d) Undefined

Q13. A bucket of water of mass $10\text{ kg}$ is pulled up from a well of depth $5\text{ m}$ at a constant speed. The work done by the lifting force is ($g = 10\text{ m/s}^2$):

(a) $500\text{ J}$

(b) $-500\text{ J}$

(c) $250\text{ J}$

(d) $50\text{ J}$

Q14. In Question 13, the work done by the gravitational force on the bucket during the lift is:

(a) $500\text{ J}$

(b) $-500\text{ J}$

(c) $0\text{ J}$

(d) $250\text{ J}$

Q15. Which of the following force-displacement angle ($\theta$) values yields maximum positive work?

(a) $\theta = 0^\circ$

(b) $\theta = 45^\circ$

(c) $\theta = 90^\circ$

(d) $\theta = 180^\circ$

Q16. A force $\vec{F} = 5\hat{k}\text{ N}$ acts on a particle while it moves from point $(0, 0, 0)\text{ m}$ to $(2, 3, 0)\text{ m}$. The work done is:

(a) $10\text{ J}$

(b) $15\text{ J}$

(c) $25\text{ J}$

(d) $0\text{ J}$

Q17. The scalar product $\vec{A} \cdot \vec{B}$ equals zero when:

(a) $\vec{A}$ and $\vec{B}$ are parallel

(b) $\vec{A}$ and $\vec{B}$ are perpendicular

(c) $\vec{A}$ and $\vec{B}$ are in opposite directions

(d) Magnitudes of $\vec{A}$ and $\vec{B}$ are equal

Q18. A body of mass $2\text{ kg}$ is held stationary at a height of $5\text{ m}$ above the ground. The work done by the holding force is:

(a) $100\text{ J}$

(b) $10\text{ J}$

(c) $0\text{ J}$

(d) $50\text{ J}$

Q19. If the magnitude of force is $10\text{ N}$, displacement is $2\text{ m}$, and work done is $-10\text{ J}$, the angle between force and displacement is:

(a) $0^\circ$

(b) $60^\circ$

(c) $120^\circ$

(d) $180^\circ$

Q20. A force $\vec{F} = (a\hat{i} + b\hat{j} + c\hat{k})$ acts on a body undergoing a displacement $\vec{s} = (\Delta x \hat{i} + \Delta y \hat{j} + \Delta z \hat{k})$. The total work done is given by:

(a) $(a\Delta x)(b\Delta y)(c\Delta z)$

(b) $a\Delta x + b\Delta y + c\Delta z$

(c) $(a + b + c)(\Delta x + \Delta y + \Delta z)$

(d) $\sqrt{(a\Delta x)^2 + (b\Delta y)^2 + (c\Delta z)^2}$

Answer Key & Brief Solutions

Q#	Answer	Brief Solution
Q1	(b)	$W = Fs \implies [N][m] = (\text{kg}\cdot\text{m/s}^2)(\text{m}) = \text{kg}\cdot\text{m}^2/\text{s}^2$.
Q2	(b)	$[W] = [\text{Force}][\text{Displacement}] = [\text{M L T}^{-2}][\text{L}] = [\text{M L}^2 \text{T}^{-2}]$.
Q3	(d)	Work is defined as $W = \vec{F} \cdot \vec{s}$, which is a scalar (dot) product.
Q4	(c)	$W = F s \cos(0^\circ) = 10 \times 5 \times 1 = 50\text{ J}$.
Q5	(b)	$W = F s \cos(60^\circ) = 20 \times 4 \times 0.5 = 40\text{ J}$.
Q6	(c)	Gravity acts downward ($\downarrow$) and displacement is horizontal ($\rightarrow$), so $\theta = 90^\circ \implies W = 0$.
Q7	(c)	Kinetic friction acts opposite to the direction of relative motion ($\theta = 180^\circ$), making work negative.
Q8	(a)	$W = \vec{F} \cdot \vec{s} = (2)(4) + (3)(-2) = 8 – 6 = 2\text{ J}$.
Q9	(a)	$\vec{s} = \vec{r}_2 – \vec{r}_1 = 3\hat{i} + 4\hat{j}$. $W = (3)(3) + (2)(4) = 9 + 8 = 17\text{ J}$.
Q10	(a)	$W = (3)(2) + (c)(-4) = 0 \implies 6 – 4c = 0 \implies c = 1.5$.
Q11	(b)	Centripetal force is radial while instantaneous displacement is tangential ($\theta = 90^\circ \implies W = 0$).
Q12	(c)	$\cos(120^\circ) = -0.5 < 0$, making work negative.
Q13	(a)	Lifting force $F = mg = 10 \times 10 = 100\text{ N}$ upward. $W = 100 \times 5 = 500\text{ J}$.
Q14	(b)	Gravity acts downward while displacement is upward ($\theta = 180^\circ$). $W_g = -(mgh) = -500\text{ J}$.
Q15	(a)	$\cos(0^\circ) = 1$ is the maximum value of cosine.
Q16	(d)	$\vec{s} = (2-0)\hat{i} + (3-0)\hat{j} + (0-0)\hat{k} = 2\hat{i} + 3\hat{j}$. $W = 5\hat{k} \cdot (2\hat{i} + 3\hat{j}) = 0\text{ J}$.
Q17	(b)	$\vec{A} \cdot \vec{B} = AB \cos(90^\circ) = 0$.
Q18	(c)	Since displacement $s = 0$, work done is zero.
Q19	(c)	$W = F s \cos\theta \implies -10 = 10 \times 2 \times \cos\theta \implies \cos\theta = -0.5 \implies \theta = 120^\circ$.
Q20	(b)	Standard vector algebraic expansion of a 3D dot product: $F_x s_x + F_y s_y + F_z s_z$.