1- A car traveling 28 m/s is traveling towards a cliff that is 65 m away when it
ID: 2300141 • Letter: 1
Question
1- A car traveling 28 m/s is traveling towards a cliff that is 65 m away when it slams on the breaks. How good do the breaks have to be (i.e. what acceleration do they need to provide) in order to stop the car before it goes over the cliff ?
2-A car traveling 28 m/s towards a cliff does not brake. it goes off the edge at that speed and proceeds to fall 250 m to the (flat) ground beneath the cliff. How far from the base of the cliff the car land ?
3-if you were to slide an 18 kg curling stone 32 m along the ice, with an intial speed of 2.1 m/s, what constant coefficient of kinetic friction would be needed?
4-you are bringing a 35 kg wheeled cart (i.e. no friction) down a 3.5 m long ramp at an angle of 18 from the horizontal. Assuming it started from rest, how much (constant) force would you have to apply on the cart to ensure that it was traveling at 1.5 m/s when it reaches the bottom of the ramp ?
5-A 20 kg box is lifted up a 10 m long frictionless slope (?/8 above the horizontal) by a 10 kg counterweight allowed to freefall. if the box starts from rest, how fast is it moving by the time it reaches the top of the slope ?
6-A 85 hockey player traveling towards the goal at 6.0 m/s is checked from the right side by 95 kg player traveling at 4.0 m/s and begin brawling. What velocity (i.e. the speed and direction) of the brawl ?
Explanation / Answer
1. by using equation of motion
v^2-u^2=2as
here we get
a=-6.03
2. range of horizontal launch is =Vo(2h/g)^(1/2)
R=28(2*250/9.81)^(1/2)
=200m
3. by equation of motion
v^2-u^2=2as
we get a=-0.068
uN=M*a
uMg=Ma
u=a/g
u=0.00693
4.
by equation of motion
v^2-u^2=2as
a=0.45 m/s^2
F-mgsin(18)=ma
F=ma+mgsin(18)
=121.85N
6. conservation of momentum
m1v1+m2v2=(m1+m2)V
v=4.971 at an angle of 33.69 degree
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.