Question

2. A force of (5+3x) N acting on a body of mass 20 kg along the x-axis displaces it from
x = 2m to x = 6m. The work done by the force is
a) 20 J
b) 48 J
c) 68 J
d) 86 J
3. A body moves from a position 1, = (21-33 - 4k) m to a position, 1 = (31 - 4j+5k) m
under the influence of a constant force F = (41 +i+6K) N. The work done by the force is
a) 57 J
b) 58 J
c) 59 J
d) 60 J
​

Answer 1

Question:

A force of (5+3 x) N acting on a body of mass 20 kg along the x-axis displaces it from  x₁ = 2 m to x₂ = 6 m. The work done by the force is _______.

Answer:

• The Work Done (W) by the force is 68 Joules.

Given:

1. The given relation is F = 5 + 3 x
2. Displacement given x₁ = 2 m to x₂ = 6 m

Explanation:

$\rule{300}{1.5}$

From the formula we Know,

$\large \bigstar\; \boxed{\tt W = \displaystyle\int\limits^{x_2}_{x_1} \tt F.dx}$

$\frak{Here}\begin{cases}\text{W Denotes Work Done}\\\text{F Denotes the variable Force}\end{cases}$

Now,

$\large \boxed{\tt W = \displaystyle\int\limits^{x_2}_{x_1} \tt F.dx}$

Substituting the values,

$\displaystyle \dashrightarrow\tt W=\displaystyle\int\limits^{x_2}_{x_1} \tt (5+3x).dx\\\\\\\dashrightarrow\tt W=\displaystyle\int\limits^{6}_{2} \tt (5+3x).dx\\\\\\\dashrightarrow\tt W=\displaystyle\int\limits^{6}_{2} \tt 5.dx+\displaystyle\int\limits^{6}_{2} \tt 3x.dx\\\\\\\dashrightarrow\tt W=\Bigg[5\;x\Bigg]^6_2+\Bigg[\dfrac{3\;x^2}{2}\Bigg]^6_2\\\\\\\dashrightarrow\tt W=\Bigg[5\;(6-2)\Bigg]+\Bigg[\dfrac{3\;(6^2-2^2)}{2}\Bigg]\\\\\\\dashrightarrow\tt W= 5\times 4+\dfrac{3\times 32}{2}$

$\dashrightarrow\tt W= 5\times 4+3\times 16\\\\\\\dashrightarrow\tt W = 20+48\\\\\\\displaystyle\dashrightarrow \large{\underline{\boxed{\red{\tt W=68\; J}}}}$

∴ The Work Done (W) by the force is 68 Joules.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Question:

A body moves from a position r₁ = (2 i - 3 j - 4 k) m to a position r₂ = (3 i -  4 j + 5 k) m  under the influence of a constant force F = (4 i + j + 6 k) N. The work done by the force is _________.

Answer:

• The Work done (W) by the force is 57 Joules.

Given:

1. r₁ = ( 2 i - 3 j - 4 k ) m
2. r₂ = ( 3 i -  4 j + 5 k ) m
3. F = (4 i + j + 6 k) N

Explanation:

Firstly finding Net Displacement ( Δ r )

Now,

⇒ Δ r = r₂ - r₁

⇒ Δ r = 3 i - 4 j + 5 k - [2 i - 3 j - 4 k]

⇒ Δ r = 3 i - 4 j + 5 k - 2 i + 3 j + 4 k]

⇒ Δ r = (3 - 2) i + (- 4 + 3) j + (5 + 4) k

⇒ Δ r = i + (- j) + 9 k

⇒ Δ r = i - j + 9 k

From the formula we know,

⇒ W = F.  Δ r

Where,

• F Denotes Force.
• W Denotes Work Done.
• Δ r Denotes Net Displacement.

Now,

⇒ W = F . Δ r

Substituting the values,

⇒ W = (4 i + j + 6 k).(i - j + 9 k)

Applying cross product,

⇒ W = (4 × 1) + (1 × - 1) + (6 × 9)

⇒ W = 4 - 1 + 54

⇒ W = -1 + 58

⇒ W = 57

⇒ W = 57 J

∴ The Work done (W) by the force is 57 Joules.

$\rule{300}{1.5}$

Answer 2

Work done by the force will be (C) 68 J.

Work done by the force will be (A) 57 J.

Explanation 1

We know, work done is the force applied on an object that produces some displacement of the body.

Mathematically, $W=F.s$

If function of force with respect to any variable is given, we integrate the force with the variable within certain limits to get the work done.

$dW=\int\limits^a_b {F} \, dx$

Here, we have,

Force = $(5+3x)N$

Mass of object = $20kg$

Limits = $(2,6)$

Substituting the given values in the equation, we get

$dW=\int\limits^6_2 {(5+3x)} \, dx$

$dW=\int\limits^6_2 {3x} \, dx +\int\limits^6_2 {5} \, dx$

$dW=3\int\limits^6_2 {x} \, dx +5\int\limits^6_2 {} \, dx$  

$dW=\frac{3}{2}x^{2} +5x$

Substituting the given limits in the equation, we get

$W=(\frac{3}{2}(6)^{2}+5(6) )-(\frac{3}{2}(2)^{2}+5(2) )$

$W=\frac{3}{2}(36-4)+(30-10)$

$W=48+20$

Hence, $W=68J(c)$

Explanation 2

We know, if force and the displacement of an object in vector form are given, the work done is given by the scalar (dot) product of both.

If the body moves from point A to point B under a force F then,

We have been given that

Force $F=(4i+j+6k)N$

Position vector at A  $(x_{1} )=(2i-3j-4k)m$

Position vector at B $(x_{2} )=(3i-4j+5k)N$

Hence, the work done is given by

$W=F$ Δ$x$    $;$ $or$

$W=F(x_{2}- x_{1} )$  

Substituting the given values in the equation, we get

$W=(4i+j+6k).((3i-4j+5k)-(2i-3j-4k))$

$W=(4i+j+6k).(3i-4j+5k-2i+3j+4k)$

$W=(4i+j+6k).(i-j+9k)$

$W=4-1+54$

Hence,  $W=57J(a)$