5. A uniform meter stick is supported as its center of mass (50cm mark) There in

Question

5. A uniform meter stick is supported as its center of mass (50cm mark) There ins suspended at the 18Scm mark You are given a 200g mass o anach so the ter torque is zero At what cm ponition.on the mse ik do you uspend this Position on meter stick 6 A horizontal spring with a spring constant of 755N/m maximum velocity of the object once the spring r of the spring and then the spring is released Assuming a friction-less surface, whas is the returns to its natural Velocity 7. If the velocity of an object were to double, what would happen to its kinetic energy? a) KE would double b) KE would not change because velocity has nothing to do with KE c) KE would decrease by a factor of two eye would increase by a factor of four 8. To achieve maximum torque, a) The applied force must be in the same direction as the moment arm b) The applied force must be as close to the pivot point as possible. c) The applied force must be directed straight toward the pivot point The applied force must be at 90 degrees to the moment arm totally bounce off of one another. What is the final velocity of the 3kg object after the collision? In which direction it is traveling? 9 A 3kg object traveling to the right at 4m/s collides with a stationary 7kg object. The two objects Direction 10. A 46 diameter wheel, rolling on the ground, makes four complete rotations what is the linear distance (tangential distance) the wheel traveled? 184- distance =

Explanation / Answer

KEmax = Umax

(1/2)*m*Vmax^2 = (1/2)*k*x^2

0.2*Vmax^2 = 755*0.15^2

Vmax = 9.22 m/s

===============

7)

Kinetic energy K1 = (1/2)*m*v1^2

K2 = (1/2)*m*v2^2

v2 = 2*v1

K2 = (1/2)*m*(2v1)^2

K2 = 4*(1/2)*m*v1^2

K2 = 4*k1

OPTION (d)

================

torque = r*F*sintheta

for torque to be maximum

sintheta value should be maximum

maximum value of sintheta = 1

sintheta = sin90

theta = 90

OPTION (d)

==========================

9)

ELASTIC COLLISION

m1 = 3 kg                       m2 = 7 kg

speeds before collision

v1i = 4 m/s                    v2i = 0 m/s

speeds after collision

v1f = ?                         v2f = ?

initial momentum before collision

Pi = m1*v1i + m2*v2i

after collision final momentum

Pf = m1*v1f + m2*v2f

from momentum conservation

total momentum is conserved

Pf = Pi

m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2

KEf =   0.5*m1*v1f^2 + 0.5*m2*v2f^2

KEi = KEf

0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)

solving 1&2

we get

v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)

v1f = ((3-7)*4 + (2*7*0))/(3+7)

v1f = -1.6 m/s

direction left

==================

10)

distance s = 4*2*pi*r = 4*2*pi*0.23 = 5.78 m

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