Calculating Minimum Stopping Distance for Truck & Box

[NAkid]

Homework Statement

The coefficient of static friction between the floor of a truck and a box resting on it is 0.39. The truck is traveling at 87.8 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?

Homework Equations

Fs=Us*N

The Attempt at a Solution

I really am not sure how to work this out. I attempted using formula Vf^=V0^2 + 2adx but I don't think that's right. Help please!

 

[NAkid]

Sorry, my other question is this:
M1 and M2 are two masses connected as shown. The pulley is light and frictionless. Find the mass M1, given that M2 (7.00kg) accelerates downwards at 3.43m/s2, that theta is 20.0o, and that muk is 0.390.

I'm pretty sure I set it up correctly:
for block 2 i said, T-mg=ma --> T=ma+mg
for block 1, T-fk-mgsin(theta)=ma AND N-mgcos(theta)=0 (no acceleration in vertical direction)

also, i know that m2>m1 for m2 to accelerate downwards

basically i plugged everything in but the answer is still wrong..

 

[anastasiaw]

Homework Statement

The coefficient of static friction between the floor of a truck and a box resting on it is 0.39. The truck is traveling at 87.8 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?

Homework Equations

Fs=Us*N

The Attempt at a Solution

I really am not sure how to work this out. I attempted using formula Vf^=V0^2 + 2adx but I don't think that's right. Help please!

I'm also having trouble with this problem. Here is what I tried (which didn't work):
First my numbers are different. Us=0.25 and v=81.4 km/hr.

First I converted 81.4 km/hr to 22.61 m/s.

Fs = Us*N => ma = Us*mg => a = Us*g => a = (0.25)(-9.81) = -2.4525 m/s^2
v^2 = v0^2 + 2a(delta x) => 0 = 22.61 + 2(-2.4525)(delta x) => delta x = 4.61 m

This is the wrong answer according to LONCAPA, so I don't really know what to do now.

 

[spizma]

First I converted 81.4 km/hr to 22.61 m/s.

Fs = Us*N => ma = Us*mg => a = Us*g => a = (0.25)(-9.81) = -2.4525 m/s^2
v^2 = v0^2 + 2a(delta x) => 0 = 22.61 + 2(-2.4525)(delta x) => delta x = 4.61 m

This is the wrong answer according to LONCAPA, so I don't really know what to do now.

You forgot to square the initial velocity in the second equation.

 

[anastasiaw]

You forgot to square the initial velocity in the second equation.

Oh simple mistakes... they kill me.

I have the right answer now... 104.2 m

 

[anastasiaw]

Sorry, my other question is this:
M1 and M2 are two masses connected as shown. The pulley is light and frictionless. Find the mass M1, given that M2 (7.00kg) accelerates downwards at 3.43m/s2, that theta is 20.0o, and that muk is 0.390.

I'm pretty sure I set it up correctly:
for block 2 i said, T-mg=ma --> T=ma+mg
for block 1, T-fk-mgsin(theta)=ma AND N-mgcos(theta)=0 (no acceleration in vertical direction)

also, i know that m2>m1 for m2 to accelerate downwards

basically i plugged everything in but the answer is still wrong..

I'm also having trouble with this one if anyone has any advice to offer.