sorry, but hw question

A 6.0kg box is dragged 4.0m along a horizontal floor with a string with tension T=30.0N. The string makes a 40 degree angle with the horizontal. The coefficient of friction between the box and the floor is uk=0.3. If the initial speed of the box is 2.0m/s, what is the final speed?

Read your damn textbook.

He lent it to a friend

want to solve for vf
v0 is known
d is known
need to find a

F = ma --> a = F/m

F = T_x - F_friction

T_x = T cos (theta)

F_friction = m * u_k

substitute and solve

Protip: F_x = ((30.0 N)(cos(40)) - ((0.3)((6.0 kg)(9.81 m/s^2) - (30.0 N)(sin(40)))))/(6.0 kg)