Jamal pulls a 150-N sled up a 28.0 degrees slope at constant speed by a force of 100 N. Near the top of the hill he releases the sled. With what acceleration does the sled go down the hill?

Answer 1

Ws = m*g = 150 N. = Weight of sled.
m = 150/g = 150/9.8 = 15.31 kg.

F1 = 150*sin28 = 70.42 N. = Force parallel to the slope.
F2 = 150*cos28 = 132.4 N. = Force perpendicular to the slope.

a = F1/m = 70.42/15.31 = 4.60 m/s^2.

Answer 2

100-150sin(28)=150/9.8(a)
a=1.93m/s^2

150sin(28)=150/9.8(a)
a=4.6m/s^2

a=4.6-1.93
a=2.67m/s^2