- #1

- 298

- 43

- Homework Statement
- A 96-kg crate, starting from rest, is pulled across a floor

with a constant horizontal force of 350 N. For the first 15 m

the floor is frictionless, and for the next 15 m the coefficient

of friction is 0.25. What is the final speed of the crate?

- Relevant Equations
- conservation of energy

so I haven't looked at the solution yet, but I know that a 100% the velocity needs to be bigger, but analytically, I get a - sign instead of a + sign as you'll see at the final square root.

So for the first 15meters of the motion all you should know is that ##v_1 = 10.458 m/s##.

for the 2nd part:

energy is conserved right. 2 NC-forces are acting on it but net force is positive however.

So ##E_1 = E_2 + W_{NC}## (1)

##W_{NC} = F{net,NC}*\Delta x##

to get

so ##F_r = \mu_kN = \mu_kmg##

##F_{net,NC} = F - \mu_kmg = 350 - 0.25*96*9.81 = 114.56N##

Now back to ##(1)##:

##\frac12mv_1^2 = \frac12mv_2^2 + F_{net,NC}\Delta x## where work done by the NC forces is pos.

rearranging to ##v_2##, we get:

##v_2 = \sqrt{\frac12mv_1^2 - F_{net,NC}\Delta x}##

##v_2 = 8.6m/s##

So for the first 15meters of the motion all you should know is that ##v_1 = 10.458 m/s##.

for the 2nd part:

energy is conserved right. 2 NC-forces are acting on it but net force is positive however.

So ##E_1 = E_2 + W_{NC}## (1)

##W_{NC} = F{net,NC}*\Delta x##

to get

**##F_{net,NC}## ,**we have ##F = 350N## and the friction force is found from an FBDso ##F_r = \mu_kN = \mu_kmg##

##F_{net,NC} = F - \mu_kmg = 350 - 0.25*96*9.81 = 114.56N##

Now back to ##(1)##:

##\frac12mv_1^2 = \frac12mv_2^2 + F_{net,NC}\Delta x## where work done by the NC forces is pos.

rearranging to ##v_2##, we get:

##v_2 = \sqrt{\frac12mv_1^2 - F_{net,NC}\Delta x}##

**WHERE i GET A - SIGN INSTEAD OF A + SIGN HERE,**__WHY?__##v_2 = 8.6m/s##