Question

A 1.5 kg box is initially at rest on a horizontal surface when at t = 0 a horizontal force $\overrightarrow { \mathrm { F } } = ( 1.8 \mathrm { t } ) \hat { \mathrm { i } } \mathrm { N }$ (with t in seconds), is applied to the box. The acceleration of the box as a function of time t is given by (take g = 10m/s²)

$\overrightarrow { \mathrm { a } }$ = 0 for 0 ≤ t ≤ 2.85

$\overrightarrow { \mathrm { a } }$ = (1.2t – 2.4) $\hat { \mathbf { i } }$ m/s² for t > 2.85

The coefficient of kinetic friction between the box and the surface is (use g = 10 m/s²) :

• A. 0.12
• B. 0.24
• C. 0.36
• D. 0.48

Answer 1

Answer: (B)

0.24

Answer 2

As motion starts after t > 2.85, then for t = 3.0 s.

As motion starts after $t > 2.85$, then for $t = 3.0 \mathrm {~s}$. $f = \mu ( 15 )$ Acceleration at $t = 3 \mathrm {~s}$ $\mathrm { a } = 1.2 \times 3 - 2.4$ $= 1.2 \mathrm {~m} / \mathrm { s } ^ { 2 }$ Applying NLM in horizontal direction $5.4 \mathrm {~N} - \mu ( 15 ) = 1.5 \times 1.2$ $\Rightarrow \quad \mu = 0.24$

Acceleration at t = 3s

a = 1.2 × 3 - 2.4 = 12 m/s²

Applying NLM in horizontal direction

5.4 N - μ(15) = 1.5 × 12 ⇒ μ = 0.24