A 5-kg package slides 1.5meters down a long ramp that is inclined at 12° below horizontal. The coefficient of kinetic friction between the package and ramp is μk=0.310 . Calculate the total work done on the package. If the package has a speed of 2.2 m/sec. at the top of the ramp, what is its speed after sliding 1.5m down the ramp???
2007-03-11 03:01:52 · 1 answers · asked by jet 3 in Science & Mathematics ➔ Physics
You need to draw a picture of what is going on. From this the following equation can be derived
F + mgsin Theta - μk *mg cos Theta = 0
where F is the driving force , m is the mass of package, g is the acceleration due to gravity, μk is the coefficient of kinetic friction, and theta is the angle of the ramp. The mgsin Theta represents component of weight acting with the driving force, and μk *mg cos Theta represent the friction acting against the driving force.
F= mg (μk cos theta - sin theta)
F= 5*9.81 (0.31 *cos 12 - sin 12) = 49.05*(0.30302257-0.2079116908) = 4.675N
Work done = Force* distance moved
work done = 4.675 *1.5 = 7.0125J
From the force we can calculate the acceleration of the package by using F=ma
a= F/m= 4.675/5= 0.935 m/s.
We can calculate the velocity at the end of the slide by
v^2 = u^2 + 2as
where v and U are the final and initial velocity of parcel respectively , acceleration, s is distance.
v^2 = 2.2^2 + 2*0.935*1.5
v^2 = 4.84 +2.805
v^2= 7.645
v= 2.76 m/s.
Hope this helps.
2007-03-11 03:43:59 · answer #1 · answered by The exclamation mark 6 · 0⤊ 0⤋