To go from this…………………………...to this.
An object experiences a force of 19.97 N for a time period of
4.58 s. What is the impulse of the object?
J=F(
t)
J= 19.97 (4.58)
For what time must you exert a force of 45 N to get an
impulse of 16 Ns?
16=45(x)
X= .36
A 142.5 kilogram motorcycle is moving at a speed of 67.5
m/s. What is the momentum of the cycle?
P=mv
P=142.5(67.5)
What is the momentum of a 23 Kg cannon shell going 530
m/s?
P=mv
P=23(530)
P= 12,190
Judy (mass=40.0 kg) is standing on slippery ice and catches
her leaping dog, Atti (mass=15 kg), moving horizontally at
3.0 m/s. What is the speed of Judy and her dog after the
catch?
mJudyi(vJudyi)+ mAtti(vAtti)= mJudyf(vJudyf )+
mAttif(vAttif )
(40)(0) + (15)(3)=
40 kg
V= 0 m/s
P= 0
P=0+45
P=45
15 kg
V= 3m/s
P=45 (15x3)
=
55 kg
V= 45=55(v)
.82 m/s
P=45
A 120 kg lineman moving west at 2 m/s tackles an 80 kg
football fullback moving east at 8 m/s. After the collision,
both players move east at 2 m/s. What is their final
momentum?
80 kg
V= 8 m/s
P= 640
P= 640+-240
=400
120 kg
V= -2 m/s
P=-240
200 kg
V=2 m/s
P= 400
Amplitude
Period
1. A force of 600 Newtons will compress a spring 0.5
meters. What is the spring constant of the spring?
2. A spring has spring constant 0.1 m/Newton. What
force is necessary to stretch the spring by 2 meters?
FHooke’s=-k( x)
1.
600= -.5 (x)
=-1200
2. F= -.1(2)
=-0.2
What is the force required to stretch a spring whose constant
value is 100 N/m by an amount of 0.50 m?
F Hooke’s =-k( x)
F= 100(.50)
F=50 N
In a hurry to catch a cab, you rush through a frictionless
swinging door and onto the sidewalk. The force you exerted
on the door was 50N, applied perpendicular to the plane of
the door. The door is 1.0m wide. Assuming that you pushed
the door at its edge, what was the torque on the swinging
door (taking the hinge as the pivot point)?
Solution
T=(1.0m) (50N)
T= 50Nm
A force of 20 N is applied perpendicular to the end of a bar of
length 0.5 m. Calculate the torque produced by the force.
T= (0.5)(20N)
T=10Nm