Engineering Mechanics example (Gate-2005)

Que: A time variation of the position of a particle in rectilinear motion is given by  x = 2t³ + t² + 2t  If v is the velocity and a is the acceleration of particle in constant units, the motion started with (Gate-2005)

(a) v = 0, a = 0

(b) v = 0, a = 2

(c) v = 2, a = 0

(d) v = 2, a = 2

Ans: (d)

Explanation: 

x = 2t³ + t² + 2t  So,

Velocity v = dx/dt = 6t² + 2t +2

Acceleration  a = 12t + 2

At the start of motion  t = 0, Then

v =  0 + 0 + 2 = 2

a = 0 + 2 = 2

Que: a car moving with uniform acceleration covers 450 m in a 5 second interval, and covers 700 m in the next 5 second interval. the acceleration of the car is (Gate-98)

(a) 7 m/s²
(b) 10 m/s²
(c) 25 m/s²
(d) 50 m/s<sup>2

Ans: (b)
Explanation: Here 
u = initial velocity
v = final velocity
a = acceleration
t = time
s = distance

using the relation</sup> s = ut + 1/2 at² and v=u + at, for the first 5 second interval,we have

450 = 5u + 25/2 a   ___(i)

v = u + 5a

for the next 5 second interval (v=u + 5a) will be the initial velocity, then

700 = (u + 5a) 5 + 25/2 a  

700 = 5u + 25a + 25/2 a   ___(ii)

From equations (i) and (ii), we get

700-450 = (5u +25a + 25/2 a) – (5u + 25/2 a)

250 = 25a

a = 250/25

a =10 m/s²

Que: How much force is required to lift the weight?

A) 40 lbs          

B) 50 lbs          

C) 60 lbs         

D) 70 lbs

Ans: (C) 
Solution: 60 lbs is needed to lift the weight. It can be calculated like this:

 f = (w x d1)/d2
 f = (80 x 9)/12
 f = (720)/12
 f = 60 lbs

Simple Machines