Question 125: Laws of Motion

Question

A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F = 20 N, making an
angle of 30o with the horizontal, as shown in the figures. The coefficient of friction between the block and
floor is $$\mu $$ = 0.2. The difference between the accelerations of the blocks, in case (B) and case (A) will be :
(g = 10 ms–2)

Accepted Answer

N = mg + 20 sin30
o
= 50 + $$20 	imes {1 \over 2}$$
= 60 N
Limition friction, f
L
= $$\mu $$N = 0.2 $$ 	imes $$ 60 = 12 N
And horizontal force(F
x
) = 20 cos30
o
= 10$$\sqrt 3 $$ = 17.32 N
As F
x
> f
L
, then the block will move with acceleration a
A
.
a
A
= $${{{F_x} - {f_L}} \over {{m_A}}}$$ = $${{17.32 - 12} \over 5}$$ = $${{5.32} \over 5}$$
N + 20 sin30
o
= mg
N = mg - 20 sin30
o
= 50 - 10 = 40 N
Limition friction, f
L
= $$\mu $$N = 0.2 $$ 	imes $$ 40 = 8 N
And horizontal force(F
x
) = 20 cos30
o
= 10$$\sqrt 3 $$ = 17.32 N
As F
x
> f
L
, then the block will move with acceleration a
B
.
a
B
= $${{{F_x} - {f_L}} \over {{m_B}}}$$ = $${{17.32 - 8} \over 5}$$ = $${{9.32} \over 5}$$
Difference between acceleration a
A
- a
B
= $${{9.32} \over 5} - {{5.32} \over 5}$$ = $${4 \over 5}$$ = 0.8 m/s
2

Answer

3.2 ms–2

Answer

0.8 ms–2

Answer

0 ms–2

Answer

0.4 ms–2

Laws of Motion

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