﻿ Work Done Worksheet | Problems & Solutions # Work Done Worksheet

Work Done Worksheet
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1.
Find the centroid of the region bounded by $y$ = $x$2, 1 ≤ $x$ ≤ 2. a. ($\frac{45}{28}$, $\frac{93}{30}$) b. ($\frac{45}{30}$, $\frac{93}{28}$) c. ($\frac{45}{14}$, $\frac{93}{30}$) d. ($\frac{45}{14}$, $\frac{93}{70}$) e. ($\frac{45}{28}$, $\frac{93}{70}$)

#### Solution:

Equation of the curve is y = x2, 1 ≤ x ≤ 2.

Area bounded by the curve y = x2, x = 1 and x = 2 is
A = 12 y dx

= 12 x2 dx = (x33)12 = 8 / 3- 1 / 3= 7 / 3

x = 1 / Aab x f(x) dx and y = 1 / A ab 1 / 2 {f(x)}2 dx
are the coordinates of the centroid of the region under y = f(x) from x = a to x = b.

x = 17312 x (x2) dx = 3 / 712 (x3) dx

= 3 / 7(x44)1 2 = 3 / 28(24 - 1) = 45 / 28

y = 17312 1 / 2(x2)2 dx

= 3 / 14(x55)1 2 = 3 / 70(25 - 1) = 93 / 70

Centroid of the region is (45 / 28, 93 / 70).

2.
Find the center of mass of the point masses $m$1 = 6, $m$2 = 8 and $m$3 = 10, lying at $x$1 = - 1, $x$2 = 2 and $x$3 = 3 on the $x$-axis respectively. a. $\frac{13}{6}$ b. $\frac{6}{13}$ c. $\frac{5}{3}$ d. $\frac{2}{3}$ e. $\frac{3}{5}$

#### Solution:

Let m1, m2, m3, . . . , mr be masses which are located at the points x1, x2, x3, . . . , xr on the x -axis. Then the moment of the masses with respect to the x -axis is
Mo = i=1rmixi and the centre of mass is x = Mom where m = i=1rmi

Mo = i=13mixi = m1x1 + m2x2 + m3x3 = 6(-1) + 8(2) + 10(3) = 40

m = i=13mi = m1 + m2 + m3 = 6 + 8 + 10 = 24

Center of mass x = Mom = 40 / 24= 5 / 3

3.
Find the center of mass of the system of point masses $m$1 = 2, $m$2 = 1 and $m$3 = 4, lying at (1, 1), (1, - 1) and (1, 0) respectively. a. ($\frac{1}{7}$, $\frac{1}{7}$) b. (1, $\frac{1}{7}$) c. (7, 7) d. (1, 1) e. ($\frac{1}{7}$, 1)

#### Solution:

Let m1, m2, m3, . . . , mr be masses which are located at the points (x1, y1), (x2, y2), (x3, y3), . . . , (xr, yr) on the plane. Then the moments of masses with respect to x - axis and y - axis are
Mx = i=1rmixi, My = i=1rmiyi respectively and the centre of mass of the system is (x, y) = (Mxm, Mym) where m = i=1rmi

Mx = i=13mixi = m1x1 + m2x2 + m3x3 = 2(1) + 1(1) + 4(1) = 7

My = i=13miyi = m1y1 + m2y2 + m3y3 = 2(1) + 1(-1) + 4(0) = 1

m = i=13mi = m1 + m2 + m3 = 2 + 1 + 4 = 7

The center of mass of the system is (x, y) = (Mxm, Mym) = (7 / 7, 1 / 7) = (1, 1 / 7)

4.
Find the center of mass of the wire of length 10 units lying on the interval [0, 10], if its density at the point $x$ is $x$2. a. $\frac{3}{4}$ b. $\frac{4}{3}$ c. $\frac{2}{15}$ d. $\frac{2}{5}$ e. $\frac{15}{2}$

#### Solution:

The center of mass of a distribution of mass with line density δ(x) on the interval [a, b] is given by
x = ab x δ(x) dx ÷ ab δ(x) dx

x = 010 x · (x2) dx ÷ 010 x2 dx

= [x44]010 / [x33]010

= 1044×3103 = 10×34 = 15 / 2

So, the center of mass is 15 / 2.

5.
The center of mass of the point masses $m$1 = 2, $m$2 = 6, $m$3 = m and $m$4 = 4, lying at $x$1 = - 4, $x$2 = - 2, $x$3 = - 1 and $x$4 = 10 respectively is 1. Find $m$. a. $\frac{1}{4}$ b. $\frac{1}{2}$ c. 1 d. 2 e. 4

#### Solution:

Let m1, m2, m3, . . . , mr be masses which are located at the points x1, x2, x3, . . . , xr on the x -axis.

Then the moment of the masses with respect to the x -axis is
Mo = i=1rmixi and the centre of mass is x = Mom, where m = i=1rmi

Mo = i=14mixi = m1x1 + m2x2 + m3x3 + m4x4 = 2(- 4) + 6(- 2) + m(-1) + 4(10) = 20 - m

m = i=14mi = m1 + m2 + m3 + m4 = 2 + 6 + m + 4 = 12 + m

Center of mass x = 1

Mom = 20 - m12 + m = 1

20 - m = 12 + m m = 4

6.
The center of mass of the system of point masses $m$1 = 3, $m$2 = 2 and $m$3 = $k$, lying at (2, 1), (3, 2) and (4, 5) respectively is (3, $\stackrel{‾}{y}$). Find $\stackrel{‾}{y}$. a. $\frac{2}{11}$ b. $\frac{11}{2}$ c. $\frac{4}{11}$ d. 1 e. $\frac{11}{4}$

#### Solution:

Let m1, m2, . . ., mr be masses which are located at the points (x1, y1), (x2, y2), . . ., (xr, yr) on the plane. Then the moments of masses with respect to x - axis and y - axis are
Mx = i=1rmixi, My = i=1rmiyi respectively and the centre of mass of the system is (x, y) = (Mxm, Mym) where m = i=1rmi

Mx = m1x1 + m2x2 + m3x3 = 3(2) + 2(3) + k(4) = 12 + 4k

m = m1 + m2 + m3 = 3 + 2 + k = 5 + k

x = Mxm = 12 + 4k5 + k

12 + 4k5 + k = 3 k = 3, so m3 = 3
[x = 3.]

My = i=13miyi = m1y1 + m2y2 + m3y3 = 3(1) + 2(2) + 3(5) = 22

m = i=13mi = m1 + m2 + m3 = 3 + 2 + 3 = 8

y = Mym = 22 / 8 = 11 / 4

7.
Find the centroid of the region bounded by $y$ = $x$, 2 ≤ $x$ ≤ 4. a. ($\frac{28}{9}$,$\frac{14}{9}$) b. ($\frac{7}{9}$,$\frac{14}{9}$) c. ($\frac{14}{9}$,$\frac{28}{9}$) d. ($\frac{28}{9}$,$\frac{28}{9}$) e. ($\frac{14}{9}$,$\frac{14}{9}$)

#### Solution:

Area of the region bounded by y = x, x = 2 and x = 4 is A = 24 y dx

= 24 x dx = (x22)24 = 422 - 222 = 8 - 2 = 6

x = 1 / 624 x(x)dx = 1 / 624 x2dx = 1 / 6(x33)24 = 43 -2318 = 56 / 18= 28 / 9
[Use x = 1 / Aab x f(x) dx.]

y = 1 / 624 1 / 2 (x)2dx = 1 / 1224 x2dx = 1 / 12(x33)24 = 43 -2336 = 56 / 36= 14 / 9
[Use y = 1 / Aab 1 / 2 {f(x)}2 dx.]

So, the centroid of the region is (28 / 9, 14 / 9).

8.
When a sugar bag is moved 10 meters with a force 100 newtons, what is the work done against resistance? a. 1000 newton-meters b. 100000 newton-meters c. 100 newton-meters d. 10 newton-meters e. 10000 newton-meters

#### Solution:

Work done against resistance = force × distance

Work done = 100 × 10 = 1000 newton-meters

So, the work done against resistance is 1000 newton-meters.

9.
When a 20 pound disc is lifted 10 feet, what is the work done against gravity? a. 200 foot-pounds b. 500 foot-pounds c. 2000 foot-pounds d. 100 foot-pounds e. 1000 foot-pounds

#### Solution:

Work done against gravity = weight × height

Work done = 20 × 10 = 200 foot-pounds

So, the work done against gravity is 200 foot-pounds.

10.
What is the work done, when a spring is extended 4 cm from its natural length with a force 4000 newtons? a. 800 newton-meters b. 40 newton-cm c. 4000 newton-meters d. 400 newton-cm e. 8000 newton-cm

#### Solution:

The force F required to compress or stretch a spring is F = kx, where k is the spring constant, x is the extension or stretch of the spring.
[Hooke's law.]

k = Fx = 40004 = 1000 newton per cm
[Evaluation of k.]

Work done = 04 F(x) dx = 04 kx dx
[Work done = ab F(x) dx.]

= (kx22)04 = 1000(x22)04 = 500 × 4 × 4 = 8000 newton-cm

So, the work done is 8000 newton-cm.