Solve all the problems below. You will be graded only on answer accuracy. There
ID: 1290337 • Letter: S
Question
Solve all the problems below. You will be graded only on answer accuracy. There is no partial credit. Box your answer on the left side by each problem number 1. A particle of weight 49.05 N is acted on my a net force of 5 N, how fast (m/s) will it be going after 5 seconds has elapsed? How far (m) will it have traveled at that point in time? 2. A tennis ball is dropped from rest at 1 yard height. How fast will it be going in mi/hr just before it hits the ground? 3. A ball of mass 1/2 kg is at the end of a rope length 1 m, and it is set into circular motion with a tangential speed of 10 m/s. What is the tension in the rope 1 the rope is essentially horizontal? 4. For the same ball and rope above, if the rope is pulled in to a length of 1/2 m as it spins and it continues to spin at the new radius, what is the tension in the rope at the new radius?Explanation / Answer
1) m = W/g = 49.05/9.8 = 5 kg
F = m*(V - U)/t
5 = 5*(V-0)/5
v = 1 m/s
-----------------
2) m*g*h = 0.5*m*v^2
v = sqrt(2gh)
v = sqrt(2*0.9144*9.8) = 18 m/s
V = 18*(1/1609)*(60*60) = 40.23 mile /hr
--------
3) T = m*v^2/r = (0.5*10*10)/1 = 50 N
-----------------------
4) m*v*r = m*v1*r1
v1 = v*r/(r/2) = 20 m/s
T = m*v1^2/r1 = (0.5*20*20)/0.5 = 400 N
---------------
5) e = V2/V1
v2 = e*v1 = 0.8*(30cos30i + 30sin30j)
v2 = 20.78 i + 12 j
v2 = 24 mph
direction = 30 degrees
---------------------
6) x = R/2 = V^2*sin(2*30)/g
50 = V^2*sin60/10
V = 24 m/s
----------------
7) a = -u*g
u = 70mph = 70*(5/18) = 19.4 m/s
v = 30mph = 30*(5/18) = 8.33 m/s
v^2 - u^2 = 2*a*s
s = ((8.33*8.33)-(19.4*19.4))/(2*0.4*9.8) = 39.15 m
---------------
8) 2T
T = 110/2 = 55 N
a = (T*cos30 - u*m*g)/m
a = 2.55 m/s^2
-------------
10) m1*u1 + m2*u2 = (m1+m2)v
v = 0.105 ft/s to the left
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.