Calculating Forces and Accelerations

[physica]

1.A 25.0 kg object is attracted towards the Earth by a force of gravity of 75.0 N. How far from the Earth's centre is it?

my answer: g=Fg/m g=75N/25kg g=3 N/kg

2. A 250 kg freezer is being pushed across the floor at a constant speed by a force of 750N. What is the coefficient of friction between freezer and floor?

my answer: Fg=mg Fg=250kg(9.8N) Fg=2450kg/N
m=Fapplied/Fg m=750N/2450kg/N m=0.31kg

3. A piece cut from a bicycle inner tube is 0.70m long when it carries a load of 25.0 N the spring constant is 300 N/m. What will the length of the piece of rubber when the load is 75.0N?

my answer: F=Kx F=300N/m / 75.0N F=4m in length

4. A 60.0kg sled is coasting with a constant speed of 10.0m/s over smooth ice. It enters a 6m stretch of rough ice where the force of friction is 120N. With what speed does the sled emerge from the rough ice?

my answer: m=Ffr/Fn m=120N/588N m=0.2kg
a=?

5. A force of 5.0N gives a mass m1 and acceleration of 8.0m/s^2 and the same force gives mass m2 an acceleration of 24.0m/s^2. What acceleration would it give the two when they are fastened together?

my answer: a=Fnet/m a=5.0N/0.833kg a=6.0m/s^2

m1=(5.0N) / (8.0m/s^2) m1=0.625kg m2=(5.0N) / (24.0m/s^2) m2=0.208k

please tell me if i messed up anywhere

 

[Galileo]

1.A 25.0 kg object is attracted towards the Earth by a force of gravity of 75.0 N. How far from the Earth's centre is it?

my answer: g=Fg/m g=75N/25kg g=3 N/kg

Hi physica.
You don't seem to have answered the question here. You are asked for a distance and end with g=3 N/kg.
g is the gravitational constant which is used on the surface of the earth.
To find the approximate distance from the center of the earth, use Newton's law of gravity:
F_g=G\frac{mM}{r^2}

2. A 250 kg freezer is being pushed across the floor at a constant speed by a force of 750N. What is the coefficient of friction between freezer and floor?

my answer: Fg=mg Fg=250kg(9.8N) Fg=2450kg/N
m=Fapplied/Fg m=750N/2450kg/N m=0.31kg

Once again. You are not asked to give a mass, but a coefficient of friction.
Use F_f=\mu N, where F_f is the frictional force and N is the normal force.

3. A piece cut from a bicycle inner tube is 0.70m long when it carries a load of 25.0 N the spring constant is 300 N/m. What will the length of the piece of rubber when the load is 75.0N?

my answer: F=Kx F=300N/m / 75.0N F=4m in length

The unit of F is not meters...
(One way to check whether you did it right is to ask yourself: "Does the answer I obtained make sense?")

Use F=k(x-x0) to find x0 from the given data (x-x0, k and F are given). Then apply it again with F=75.0 N to find x-x0 for the second case.

 

[physica]

are the rest right?

 

[Galileo]

4. A 60.0kg sled is coasting with a constant speed of 10.0m/s over smooth ice. It enters a 6m stretch of rough ice where the force of friction is 120N. With what speed does the sled emerge from the rough ice?

my answer: m=Ffr/Fn m=120N/588N m=0.2kg
a=?

There are different ways to solve this one. I'd use an energy approach.
Since m and v at the beginning are given, you can calculate the kinetic energy 1/2mv^2.
The friction does negative work on the sled: W=Fd.
So you from this you know the kinetic energy at the end. you can get v from this.

5. A force of 5.0N gives a mass m1 and acceleration of 8.0m/s^2 and the same force gives mass m2 an acceleration of 24.0m/s^2. What acceleration would it give the two when they are fastened together?

my answer: a=Fnet/m a=5.0N/0.833kg a=6.0m/s^2

m1=(5.0N) / (8.0m/s^2) m1=0.625kg m2=(5.0N) / (24.0m/s^2) m2=0.208k

That's correct