1. A person pushes a 13.5-kg crate at constant speed with a force F = 75.0 N dir
ID: 1773075 • Letter: 1
Question
1. A person pushes a 13.5-kg crate at constant speed with a force F = 75.0 N directed at an angle of 45.0° to the horizontal.
(a) Calculate the friction force acting on the crate.
(b) Calculate the normal force acting on the crate.
(c) What constant force at the same angle must the person exert on the crate to accelerate it from rest to 1.50 m/s in 2.00 s?
2. A man on a sled with a total mass of 125 kg starts from rest and slides down a snow-covered slope which makes a constant angle of 35.0° with the horizontal. Assume that the slope is 75.0 m long and the coefficient of kinetic friction between the sled and the snow is 0.0900.
(a) Find the acceleration of the sled.
(b) Find the speed of the sled at the bottom of the slope.
(c) Find the work done by friction during the 75.0 m slide down the slope.
Explanation / Answer
1.
a)
frictional force is being balanced by the horizontal component of the applied force
hence
f = F Cos45
f = 75 Cos45 = 53.03 N
b)
force equation in vertical direction is given as
Fn + F Sin45 = mg
Fn + 75 Sin45 = (13.5) (9.8)
Fn = 79.3 N
c)
F' = new applied force
a = acceleration = (1.50 - 0)/2 = 0.75 m/s2
F' Cos45 - f = ma
F' Cos45 - 53.03 = 13.5 x 0.75
F' = 89.3 N
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.