Question #98429
A child standing on the floor of a balcony of a multi storey building, throws a ball straight upwards with an initial velocity of 10m/s. Assume the the floor is 25m high from the ground. Take g=9.8m/s/s.find the
a) maximum height above the ground reached by the ball.
B.) velocity with which the ball strikes the ground.

2.
A 10kg trunk lies on a horizontal rough floor,the coefficient of friction between the trunk and the floor is 0.75. Calculate magnitude of force P which is necessary to pull the trunk horizontally if the P is applied horizontally and at 30 degrees above the horizontal.

3.
A particle of mass 300g is attached to one end of a light inextensible string of length 40cm, the other end of the string being fixed at 0 on a smooth horizontal surface. If the particle describe a circle centre o, find the tension in the string when
A)The speed of the particle is 2.83m/s.
B)The angular speed of the particle is 5rad/s.
1
Expert's answer
2019-11-12T17:25:57-0500

1a) The maximum height would be


hmax=hbalcony+vi22g=25+5.1=30.1 m.h_{max}=h_{balcony}+\frac{v_i^2}{2g}=25+5.1=30.1\text{ m}.

1b) The velocity will be


vf=2ghmax=24.3 m/s.v_f=\sqrt{2gh_{max}}=24.3\text{ m/s}.

2) To make the trunk move, one needs to apply a horizontal force equal to the force of friction:


Px=μmg,P_x=\mu mg,

where


Px=P cosθ,P_x=P\text{ cos}\theta,

therefore,


P=μmgcosθ=84.9 N.P=\frac{\mu mg}{\text{cos}\theta}=84.9\text{ N}.

3a) The centripetal force acting on the particle will be equal to the tension:


T=mv2R=6 N.T=\frac{mv^2}{R}=6\text{ N}.

3b) Use the same expression, but now


v=ωR:v=\omega R:

T=mω2R=3 N.T=m\omega^2R=3\text{ N}.


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