Question

A mass of 500kg stands on a ramp inclined at 24degrees to the horizontal. The coefficient of friction
between the surfaces of the mass and ramp is 0.27. If the mass is allowed to move from rest, find the
velocity of the mass after six seconds. Assume that g = 9.8 m/sec^2. Your answer should be correct to one decimal.​

Answer 1

The net force acting on a mass $\sf{m}$ placed on a rough inclined plane of angle of inclination $\theta$ and coefficient of friction $\mu,$ along the plane is,

$\sf{\longrightarrow F=mg(\sin\theta-\mu\cos\theta)}$

Hence the net acceleration is,

$\sf{\longrightarrow a=g(\sin\theta-\mu\cos\theta)}$

Since $\sf{a=\dfrac{dv}{dt},}$

$\sf{\longrightarrow \dfrac{dv}{dt}=g(\sin\theta-\mu\cos\theta)}$

$\sf{\longrightarrow dv=g(\sin\theta-\mu\cos\theta)\ dt}$

Integrating as follows ($\sf{g,\ \theta,\ \mu}$ are independent of $\sf{t}$)

$\displaystyle\sf{\longrightarrow\int\limits_u^vdv=g(\sin\theta-\mu\cos\theta)\int\limits_0^tdt}$

$\displaystyle\sf{\longrightarrow v-u=g(\sin\theta-\mu\cos\theta)(t-0)}$

$\displaystyle\sf{\longrightarrow v=u+gt(\sin\theta-\mu\cos\theta)}$

In the question,

• $\sf{u=0\ m\,s^{-1}}$

• $\sf{g=9.8\ m\,s^{-2}}$

• $\sf{t=6\ s$

• $\sf{\mu=0.27}$

• $\sf{\theta=24^o}$

Then,

$\displaystyle\sf{\longrightarrow v=0+9.8\times6(\sin24^o-0.27\cos24^o)}$

$\sf{\longrightarrow\underline{\underline{v=9.4\ m\,s^{-1}}}}$

Answer 2

Given :-

• u ( starting velocity ) = 0 m/s²
• t = 6 sec
• ∅ = 24°
• t = 6 sec
• μ = 0.27
• g = 9.8 m/s²

Using formula :-

• V = U + gt ( sin ∅ - μ cos ∅ )

Required Answer :-

$\tt {↝\: \: v = u + gt(sin \theta - μcos \theta)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt {↝\: \: v = 0 + 9.8 \times 6(sin \: 24 - 0.27 \times cos \: 24) } \\ \\ \tt {↝\: \: v = 58.8(0.4067.. - 0.27 \times 0.9134..) } \\ \\ \tt {↝\: \: v = 58.8( 0.160081) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt {↝\: \: v = 9.412 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

more information :-

$\tt \blue {↝\: \: acceleration = \frac{dv}{dt} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \tt \blue {↝\: \: we \: \: have \: \: formula \: \: of \: \: acceleration \: \: is } \\ \\ \tt \blue {↝\: \: a = g(sin \theta - μcos \theta )} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt \blue {↝\: \: \frac{dv}{dt} = g(sin \theta - μcos \theta )\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:} \\ \\ \tt \blue {↝\: \:dv =g(sin \theta - μcos \theta )dt\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \tt \blue {↝ \:integrating \: \: from \: \: both \: \: side \: \: then \: \: we \: \: get } \\ \\ \tt \blue {↝\: \:v - u = (t - 0)g(sin \theta - μcos \theta )} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\\tt \blue {↝\: \:hence \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \tt \blue {↝\: \:v = u + gt(sin \theta - μcos \theta)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

I hope it helps you ❤️✔️