Question

A block of mass 10 kg is dragged across a horizontal surface by pulling the block with force of 100 N. The force is applied by attaching a chord to the block The chord is inclined at an angle of 60° with the horizontal If the coefficient of kinetic friction is 0.2 what is the acceleration of the block? (See Figure).

Answer 1

From the Figure,

Vertical component of the Force = F Sin 60°

= 100 × √3/2 = 50√3 N.

Now, the Normal Reaction also acts on the object in upwards direction.

∴ N + 50√3 = mg

⇒ N + 50√3 = 10 × 10

∴ N = 100 - 50√3

⇒ N = 13.39 N.

Now, Friction = Coefficient of kinetic Friction × N = 0.2 × 13.39 = 2.68 N.

Since, Friction opposes the Relative Motion,

∴ It will acts in opposite direction of motion.

Horizontal component of the Force = F Cos 60° = 100 × 1/2 = 50 N.

Now, Horizontal component is greater than the Frictional force therefore object will move in the direction of force.

∴ Force = ma = 10a,  where a is the acceleration.

∴  50 - 2.68 = 10a

⇒  10a = 47.32

∴ a = 4.73 m/s².

Hope it helps.

Answer 2

see figure to understand free body daigram of given system.

due to applied force, block has tendency to move forward so, friction acts backward as shown in figure.

body is rest in vertical direction,
so, upward forces = downward forces
Fsin60° + N = mg [ see figure, ]
N = mg - Fsin60° ..........(i)

body is moving in horizontal direction,
so, forward forces - backward forces = ma
here a is acceleration.
Fcos60° - fr = ma ........(ii)

but we know, fr = μN
from equation (i),
fr = μ(mg - Fsin60°) , put it in equation (ii),

so, Fcos60° - μ(mg - Fsin60°) = ma
now, put F = 100 N, m = 10kg , g = 10m/s² and μ = 0.2

100 × 1/2 - 0.2(10× 10 - 100 × √3/2) = 10a
50 - 0.2(100 - 50√3) = 10a
50 - 20 + 10√3 = 10a
30 + 10√3 = 10a
a = (3 + √3) m/s²

hence, acceleration = (3 + √3) m/s²