1. A tension force of 3.25 N acts horizontally on a 5.00 kg block. The block accelerates at 0.890 m/s2. What is the force of kinetic friction?

2. The 60 kg student on skates pushes a 55 kg student on skates. The 55 kg student accelerates at 2.5 m/s2. What is the acceleration of the 60 kg student?

3. A 0.160 kg ice puck collides with a 0.180 kg ice puck. Both were originally moving at the same speed. Which one has the larger change in velocity?

4.A 6.0 kg block of ice rests on a surface that is tilted by θ=12°. Find the acceleration of the block, ignoring any friction.

jjennylove Dec 10, 2019

#1**0 **

1. Remember f= ma ?

the force net on the object is 3.25 one direction and kinetic frictional force in the OPPOSITE direction

(3.25 - K) = 5 kg .890 m/s^{2 }Solve for K

2. Same thing here f=ma

I THINK what they mean is 'equal and opposite reactions' and conservation of energy etc

f = ma for the 55 kg student

= 55 kg (2.5 m/s^{2) }This force will also act BUT in the opposite direction for the 60 kg student

55 (2.5) = 60 kg a Solve for a

3. conservation of momentum

Think about a large truck hitting a smaller car....which will have the biggest change in velocity ???

4. You must find the force acting DOWN the plane (NOT the NORMAL force.....we are ignoring friction))

then f = ma

ElectricPavlov Dec 11, 2019

#3**0 **

1 (3.25 - K) = 5 kg .890 m/s2 Solve for K

3.25 - k = 5 (.890)

3.25-k = 4.45

k = -1.2 N

2 yes

3 yes

4 mg sin 12 is the force down-plane

f= mg sin 12 = ma

g sin 12 = a =2.04 m/s^{2} (this should make you realize that the acceleration is not dependent on the mass of the object)

ElectricPavlov Dec 11, 2019

#4**0 **

im still kind of unsure how you solved the last one (4)? what did you multiply or divide?

jjennylove
Dec 11, 2019

#5**0 **

mg sin 12 is the force down-plane

f= mg sin 12 = ma (divide both sides by 'm')

g sin 12 = a

9.81 * sin 12 = a

9.81 * .20791 =a

a =2.04 m/s2

ElectricPavlov
Dec 11, 2019