Question

A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s^2. The free-body diagram shows the forces acting on the cart.

A free body diagram with 3 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline. The second vector is pointing up, labeled F Subscript N Baseline. The third vector is pointing up to the right at an angle of 35 degrees, labeled F Subscript p Baseline = 250 N. The up and down vectors are the same length.


The mass of the cart, to the nearest whole number, is

Answer 1

Split up the acceleration into horizontal and vertical components:

a = (1.4 m/s²) (cos(35º) i + sin(35º) j)

a ≈ (1.15 m/s²) i + (0.803 m/s²) j

Do the same with the pulling force (denoted by p):

p = (250 N) (cos(35º) i + sin(35º) j)

p ≈ (205 N) i + (143 N) j

By Newton's second law, the net horizontal force is proportional to the horizontal acceleration according to

∑ F = m a

The only horizontal force acting on the cart is the horizontal component of the pulling force. So we have

205 N = m (1.15 m/s²)

m ≈ 178.571 kg ≈ 180 kg