Please show your work to receive credit. Car 1 has 7 times the mass of car 2, bu
ID: 2138370 • Letter: P
Question
Please show your work to receive credit.
Car 1 has 7 times the mass of car 2, but only 0.6 as much kinetic energy. What is the ratio between the speeds of the two cars (v1/v2)? You must type your answer as a numerical value. A car of mass 1550 kg traveling at 26 m/s is at the foot of a hill that rises 120 m in 3 km. At the top of the hill, the speed of the car is 12 m/s. Find the average power delivered by the car's engine, neglecting any frictional losses. A roller coaster reaches the top of the steepest hill with a speed of 6.4 km/h. It then descends the hill, which is at an average angle of 20 degree and is 40 m long. What will its speed be when it reaches the bottom? Assume mu k = 0.12. The 2 kg block in the figure slides down a frictionless curved ramp, starting from rest at a height of h = 4 m. The block then slides d= 10 m on a rough horizontal surface before coming to rest. What is the coefficient of friction between the block and the horizontal surface? A 8 kg sled is initially at rest on a horizontal road. The sled is pulled a distance of 3.2 m by a force of 36 N applied to the sled at an angle of 30 degree to the horizontal. Find the change in the kinetic energy of the sled. After hitting a long fly ball that goes over the right fielders head and lands in the outfield, the batter decides to keep going past second base and try for third base. The 68 kg player begins sliding 2.40 m from the base with a speed of 4.49 m/s. If the player comes to rest at third base, What was the coefficient of kinetic friction between the player and the ground? A record was set for stair climbing when a man ran up the 2200 steps of the EmpireExplanation / Answer
1) m1=7*m2 gives m2/m1=1/7
and .5*m1*v1^2=.5*m2*v2^2*(.6) thus v1/v2=sqrt(.6*m2/m1)=.29
2)time taken to climb the hill= 3*1000/26=115.38 seconds
potential energy gained by car=m*g*H=1550*9.8*120
Power=energy/time=15797.6 watts
3)Friction force=coefficient*m*g*cos(theta)=.1128*m
work done against fricttion=.1128*m*d=.1128*m*40=4.51*m
potential energy lost by block=m*g*d*sin(theta)=134.072*m Joule
initial velocity= 6.4*1000/3600 m/s=1.778 m/s
let final velocity be v2:then using work energy theorem:
.5*m*(1.778)^2+134.072*m-4.51*m=.5*m*v2^2 gives v2=16.19 m/s=58.3 Km/hr
4)work done against friction= gain in potential energy
m*g*coefficient*distance on horizontal=m*g*H gives
coefficient=H/d=4/10=.4
5)Change in kinetic energy= work donw by the force= F*cos(theta)*distance=99.76 Joule
6)work done by friction force=reduction in kinetic energy
m*g*coefficient*distance=.5*m*v^2
gives coefficient=.428