- #1
decentfellow
- 130
- 1
Homework Statement
A shot putter with a mass of ##80## kg pushes the iron ball of mass of ##6## kg from a standing position accelerating it uniformly from rest at an angle of ##45^{\circ}## with the horizontal during an interval of ##0.1## seconds. The ball leaves his hand when it is ##2\text{m}## high above the level ground and hits the ground ##2## seconds later.
The minimum value of coefficient of friction if the shot putter does not slip during the shot is closest to:-
a) ##0.28##
b) ##0.38##
c) ##0.48##
d) ##0.58##
Homework Equations
##\Delta S= u_yt-\dfrac{gt^2}{2} \tag{1}##
##f_{max}=\mu_{min}N\tag{2}##
The Attempt at a Solution
First of all we need to find the acceleration of the ball. For that we use the fact that we are given with the time for which a force was applied in the ball which is ##0.1## ##\text{sec}##, the time of flight of the ball and the height of projection of the ball.
So, we have
- Time for which a force had been applied on the ball ##=0.1## ##\text{seconds}##
- Height of projection of the iron ball ##=2## ##\text{m}##
- Angle of projection ##=45^{\circ}##
- Time of flight of the iron ball ##=2## ##\text{seconds}##
From ##\text{eq}^{\text{n}}## ##(1)##, we get
$$\Delta S= u_yt-\dfrac{gt^2}{2}\implies -2=\left[(a\times 0.1)\times \sin{45^{\circ}}\times 2\right]-(5\times 4)\implies a=90\sqrt2\text{m/s}^2$$
So, the force exerted by the iron ball on the shotputter is ##F=6\times 90\sqrt2=540\sqrt2##. Now consider the FBD of the shotputter as shown below:-
From the FBD we get,
##N=F\sin\theta + Mg##
##f_s=F\cos\theta\implies \mu_{min}N=F\cos\theta\implies \mu_{min}=\dfrac{F\cos\theta}{Mg+F\sin\theta}=\dfrac{540}{800+540}=\dfrac{27}{67}\approx 0.40##
Though the question asks for the nearest value and 0.40 is nearest to 0.38, so it should be the answer and so is the case. But I am not sure if I am doing everything correct or not because my answer has a difference of 0.02 which is a considerable difference considering the options that are given.