Question 148: 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 30
o
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
)
![](https://imagex.cdn.examgoal.net/jf762wkjxv35uis/673373d6-5b77-484e-8976-3aeb901f828b/d852a840-a1e1-11e9-902e-23ffedc9804e/file-jf762wkjxv35uit.png)

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
![](['https://imagex.cdn.examgoal.net/jf762wkjxvdtc0z/52d08ac0-1c79-4f52-a567-11bb0eed1299/80bf4730-a20b-11e9-902e-23ffedc9804e/file-jf762wkjxvdtc10.png', 'https://imagex.cdn.examgoal.net/jf762wkjxvezx0x/5dce3c9b-f357-4ab5-8a8b-7f3e7ce3d19e/20f3b610-a210-11e9-902e-23ffedc9804e/file-jf762wkjxvezx0y.png'])

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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