1. A baseball thrown straight upward returns to the ground 5:0 s later. What was

Question

1. A baseball thrown straight upward returns to the ground 5:0 s later. What was its initial upward velocity? (Assume it was released at ground level.) a. 24:5 m=s b. 14:5 m=s c. 6:5 m=s d. 38:7 m=s e. 17:7 m=s

2. A ball is thrown horizontally from the top of a building at a speed of 16:5 m=s and lands 37:0 m from the building. How tall is the building? a. 41:2 m b. 24:6 m c. 17:0 m d. 65:5 m e. 38:6 m

3. A crane is lifting a 2000 kg load with a cable whose breaking strength is 22; 000 N. What is the maximum upward acceleration that the load can be given? a. 1:2 m=s2 b. 0:71 m=s2 c. 2:0 m=s2 d. 1:6 m=s2 e. 2:4 m=s2

4. A box starts from rest and slides down an incline that is 10 m long. At the bottom it is traveling at a speed of 5:0 m=s. If the incline makes an angle of 200 with respect to the horizontal, what is the coe¢ cient of friction between box and incline? a. 0:15 b. 0:23 c. 0:31 d. 0:40 e. 0:37

Please show all work for the answers, greatly appreciated

Explanation / Answer

1) given that time of flight is T = 2*u/g = 5

initial speed is u = 5*g/2 = 2.5*9.81 24.5 m/s

2) range R = u*sqrt(2h/g)

R = 37 m

u = 16.5 m/s

g =9.8 m/s^2

then h = (R^2/u^2)*(g/2) = [(37*37)/(16.5*16.5)]*(9.81/2) = 24.6 m

3) writing equation of motion

T-mg = ma

22000-(2000*9.81) = 2000*a

accelaration a = 1.2 m/s^2

4) net force m*g*sin(20) - (mu_k*m*g*cos(20)) = m*a

Work done by the net force = change in KE = Kf-Ki = Kf = 0.5*m*v^2 = 0.5*m*5*5 = 12.5*m

[m*g*sin(20) - (mu_k*m*g*cos(20))]*S = 12.5*m

9.81*sin(20) - (mu_k*9.81*cos(20)) = 12.5/10 = 1.25

(mu_k*9.81*cos(20)) = 2.105

mu_k = 0.23

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