A trailer of mass 1000 kg is towed by means of a rope attached to a car moving at a steady speed along a level road. The tension in the rope is 400 N. The car starts to accelerate steadily. If the tension in the rope is now 1650 N,with what acceleration is the trailer moving ?

1.75 ms^-2
0.75 ms^-2
2.5 ms^-2
1.25 ms^-2

Solution

Here, the force of friction is 400N.
F_net= (1650-400)=1250N

A weight W rests on a rough horizontal plane. If the angle of friction be θ, the least force that will move the body along the plane will be

W cosθ
W cotθ
W tanθ
W sinθ

Solution

f=μ W
f=W tanθ[∴μ=tan θ]

A body of mass 1 kg moving with a uniform velocity of 11ms^-1. If the value of g is 25ms^-2, then the force acting on the friction less horizontal surface on which the body is moving is

5 N
1 N
0 N
10 N

Solution

Weight of body = m g = 5 N

A ball of mass m is thrown vertically upwards. What is the rate at which the momentum of the ball changes?

Zero
mg
Infinity
Data is not sufficient.

Solution

The time rate of change of momentum is force.

A rider on a horse back falls forward when the horse suddenly stops. This is due to

inertia of horse
inertia of rider
large weight of the horse
losing of the balance

Solution

Inertia is resistance to change.

A force \(\dot{F}=8\hat{i}-6\hat{j}-10\hat{k}\) rnewton produces an accelerationof 1 ms^-2 in a body. The mass of the body is

10 kg
10\(\sqrt{2}\) kg
10\(\sqrt{3}\) kg
200 kg

Solution

\(m=\frac{\sqrt{\sqrt{8^{2}+(-6)^{2}+(-10)^{2}}}}{1}-10\sqrt{2}\) kg

The coefficient of friction between two surfaces is 0.2. The angle of friction is

sin^-1(0.2)
cos^-1(0.2)
tan^-1(0.1)
cot^-1(5)

Solution

Angle of friction = tan^-1 µ

A 0.1 kg block suspended from a massless string is moved first vertically up with an acceleration of 5ms^-2 and then moved vertically down with an acceleration of 5ms^-2. If T₁ and T₂ are the respective tensions in the two cases,then(a)(b)(c)(d)

T₂>T₁
T₁T₂= 1 N, if g=10ms^-2
T₁-T₂=1kgf
T₁T₂=9.8N,if g=9.8ms^-2

Solution

T₁=m(g+a)= 0.1(10+5)1.5N
T₂m(g-a)=0.1(10-5)0.5N
⇒T₁-T₂=(1.5-0.5)N=1N

A 4000 kg lift is accelerating upwards. The tension in the supporting cable is 48000 N. If g=10ms^-2 then the acceleration of the lift is

1 ms^-2
2 ms^-2
4 ms^-2
6 ms^-2

Solution

T = m(g+a)

∴ 48000 = 4000(10+a)

⇒ a = 2ms^-2

A mass is hanging on a spring balance which is kept in a lift.The lift ascends. The spring balance will show inits readings

an increase
a decrease
no change
achange depending on its velocity

Solution

Let acceleration of lift=a and let reactionat spring balance = R

Applying Newton’s law
R – mg =ma ⇒ R=m(g+a)
thus net weight increases,So reading of spring balance increases.

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