weight = 7 * 9.81 Newtons
force down slope = weight * sin 30
= (1/2)weight
force up slope = 381 N
net force up slope
= 381 -(1/2) weight
= 347 N
a = F/m = 347/7 = 49.5 m/s^2
v = 5 + a t
v = 5 + 49.5 t
distance = average speed * t
9 = (1/2)(v+5) t
18 = (10+49t) t
49 t^2 + 10 t - 18 = 0
t = [ -10 +/-sqrt(100+3528) ]/98
t = [-10 + 60.2 ]/98
t = .513 second
v = 5 + 49.5(.513) = 30.4 m/s
a box of mass 7 kg is pulled up a frictionless 30 degree incline by a force of 381 N which acts parallel to the incline surface. the box has an initial speed of 5m/s and is pulled 9m. what is the speed of the box at the end of the 9m.
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