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physics question please help?
A metal block of mass 1.5 kg is placed on a smooth inclined slope 40 dgrees. The block is released and slides to the bottom of the ramp, a distance of 0.75 m.
Assume an average frictional force (F) of 4.0 N along the slope during the slide.
g = 9.81 m s–2.
Calculate the speed of the block as it reaches the bottom of the slope.
1 Answer
- oldschoolLv 75 years ago
Two ways to do this.
Fnet = 1.5gsin40 - F = 9.45-4 =5.45N
Fnet = m*a -----> F/m = a = 5.45/1.5 = 10.9/3 = 3.63m/s²
Vf² = Vi² + 2*a*d = 0² + 2*3.63*0.75 = 5.45m/s
Vf = √5.45 = 2.33m/s <------
OR use conservation of energy
m*g*h = 1.5*9.8*(0.75*sin40) = 7.09 J
7.09 = KE + Energy lost to Friction = F*0.75m = mV²/2 + 4*0.75
-----> V = √2*(7.09-3)/1.5] = 2.33m/s <----- same answer
