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thebestrc
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[SOLVED] Mechanics question-Momentum&F=ur
An ice hockey player is required to strike a puck having a mass of 1.2 kg, so that it is
accelerated uniformly from rest across a smooth ice rink surface so as to cover a distance of 25 m in 8 s.
The puck travels another 8 m before coming to a rest. Determine (a) the
acceleration that the puck has to experience, (b) the minimum force the ice hockey player has to apply to the puck to achieve this acceleration, (c) the momentum of the puck at the end of the 8 s period, (d) the frictional force that brings the puck to rest, and (e) the coefficient of kinetic friction between the ice and the puck.
[tex]s=ut+1/2at^2[/tex]
[tex]F=\mu R[/tex]
[tex]f=ma[/tex]
(a)use the formula shown above u=0 t=8s s=25m therefore simply substitue into equation a=-0.78125
(b)use 3rd equation m=1.2kg a=-0.78125 f=0.9375N
(c)momentum is g=mv . Once again substitute values into equation. m=1.2kg v=6.25m/s found out by using s=1/2 (u+v)t g=7.5kg m/s
(d)&(e). Both use the 2nd formula but I'm unsure as how to go about it. I know R=mg. but F can't be the same as in f=ma can it?
Also check over the other stuff too please for any mistakes thanks.
Homework Statement
An ice hockey player is required to strike a puck having a mass of 1.2 kg, so that it is
accelerated uniformly from rest across a smooth ice rink surface so as to cover a distance of 25 m in 8 s.
The puck travels another 8 m before coming to a rest. Determine (a) the
acceleration that the puck has to experience, (b) the minimum force the ice hockey player has to apply to the puck to achieve this acceleration, (c) the momentum of the puck at the end of the 8 s period, (d) the frictional force that brings the puck to rest, and (e) the coefficient of kinetic friction between the ice and the puck.
Homework Equations
[tex]s=ut+1/2at^2[/tex]
[tex]F=\mu R[/tex]
[tex]f=ma[/tex]
The Attempt at a Solution
(a)use the formula shown above u=0 t=8s s=25m therefore simply substitue into equation a=-0.78125
(b)use 3rd equation m=1.2kg a=-0.78125 f=0.9375N
(c)momentum is g=mv . Once again substitute values into equation. m=1.2kg v=6.25m/s found out by using s=1/2 (u+v)t g=7.5kg m/s
(d)&(e). Both use the 2nd formula but I'm unsure as how to go about it. I know R=mg. but F can't be the same as in f=ma can it?
Also check over the other stuff too please for any mistakes thanks.