A crate (14 kg) is pulled by a force (parallel to the incline) up a rough incline (25 degrees). The crate has an initial speed of 1.56 m/s. The crate is pulled a distance of 7.99m on the incline by a 150N force.
mu(greek letter)=0.285
The acceleration of gracity is 9.8 m/s^2
a) What is the change in kinetic energy of the crate? Answer in units of J.
b) What is the speed of the crate after it is pulled the 7.99m? Answer in units of m/s.
2006-11-02 12:33:36 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Given: mass, m, of crate=14kg
therefore, weight of crate=14g
initial speed of crate=1.56m/s
applied force along the incline=150N
distance traversed by crate=7.99m
mu=0.28
acceleration of gravity,g=9.8m/s^2
angle of incline=25 degree
a) Initial kinetic energy, KE=1/2mv^2
=1/2*14*1.56^2=17 Joules
force of friction=mu*R,
where R=14*9.8cos25=124.3N
Net force,F=150-124.3
=25.7N
F=ma
25.7=14a
a=25.7/14
=1.84m/s^2
v^2-u^2=2as
v^2-1.56^2=2*1.84*7.99
v^2=29.4+2.43
=31.8
v=5.64m/s
Final KE=1/2*14*5.64^2
=222.6 Joules
Change in KE=222.6-17
=205.6 Joules
b) 5.64m/s as computed in a) above.
2006-11-04 02:37:55 · answer #1 · answered by tul b 3 · 0⤊ 1⤋
I think you should do your own homework... so I can back to mine...
Oh, man, I can't figure it out. When you say J, do you mean Joules?
2006-11-02 20:45:35 · answer #2 · answered by dirtywienerdog 1 · 0⤊ 1⤋