+1993 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.
+36916 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/s2 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/s2) 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
+36916 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/s2 (this should make you realize that the acceleration is not dependent on the mass of the object)
+1993 im still kind of unsure how you solved the last one (4)? what did you multiply or divide?
+36916 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
