Question

Bob’s 90.0 kg apple cart accidently started rolling down a hill. When Bob caught up with the cart, it was travelling at 2.00 m/s. Bob applied a braking force (parallel to the hill) of 400 N (force F on the diagram) over a distance of 3.50 m. If you ignore air resistance and friction, how fast was the cart going after this distance? The angle from the groud to the begining of the hill is 25 degrees

Answer 1

mass of th apple cart m=90kg

initial speed of the cart is v1=2 m/sec

final speed of the cart is v2

distance travelled, d=3.5 m

braking force, F=400 N

angle, theta=25 degrees,

by using energy relation,

1/2*m*v2^2=1/2*m*v1^+m*g*h-F*d

1/2*m*v2^2=1/2*m*v1^+m*g*d*sin(theta)-F*d

v2^2=v2^+2*g*d*sin(theta)-(2*F*d)/m

v2^2=2^2+(2*9.8*3.5*sin(25)-(2*400*3.5)/(90)

===> v2=1.37 m/sec

velocity of the cart after travelling distance d=3.5m is v2=1.37 m/sec