Question

A 10 kg mass is resting on a horizontal surface and horizontal force of 80 N is applied. If μ = 0.2, the ratio of acceleration without friction and with friction is (g = 10ms−2)

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Answer 1

Explanation:

a

2

a

1

=

m

F−μmg

m

F

a

1

=

m

F

=

10

80

=8 m/s

2

a

2

=

m

F−μmg

=

10

80−0.2×10×10

=6 m/s

2

∴

a

2

a

1

=

6

8

=

3

4

Answer 2

Answer:

• $\bf\purple{\frac{4}{3}}$

Explanation:

Given:

• A 10 kg mass is resting on a horizontal surface and horizontal force of 80 N is applied.

To Find:

• If μ = 0.2, the ratio of acceleration without friction and with friction is (g = 10ms−2)

Solution:

$\implies\:\:\frac{a_{1}}{a_{2}} = \frac{\frac{F}{m} }{ \frac{F - \mu mg}{m} }$

$\implies\:\:a_{1} = \frac{F}{m} = \frac{80}{10} = 8m/s^2$

$\implies \:\:a_{2} = \frac{F - \mu mg}{m} = \frac{80-0.2 \times 10 \times 10}{10} = 6m/s^2$

$\implies\:\:\frac{a_{1}}{a_{2}} = \frac{8}{6} = {\red{\bf{\frac{4}{3}}}}$