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Chapter 34: Practice Exercises

- M and m are hung over a massless pulley.

- At the bottom of the circle, the mass is observed to have a speed of 10 m/s.

- A car and driver have a lot of weight.
- The car goes over the top of the hill at a speed of 5 m/s.

- A hockey puck with a mass of less than one kilo is sliding along the ice.
- The puck's speed is 20 m/s.
- The floor has a 0.35 coefficient of friction, which is equal to the puck crossing over it.

- The natural length of the spring is 0.45 m and it is attached to the top of the incline.
- The incline is long.
- The spring is stretched down the incline because of a mass attached to it.

- The two blocks are being pushed along the floor by a 20-N force.

- A force of 20 N acts horizontally on a mass of 10 kg being pushed along a frictionless incline that makes a 30 degree angle with the horizontal, as shown below.

- A mass is released from rest on an incline that makes a 42 degree angle with the horizontal.

- A block is placed on top of another block.

- The block has a 0.35 coefficient of friction.
- There is a force horizontal F on the block.

- A road that is banked at an angle is not necessary for a car to stay on the road.
- A car is traveling at a speed of 25 m/s and the road has a radius of 40 m.

- The "rotor" is an amusement park ride that can be modeled as a rotating cylinder.
- A person is held against the sides of the ride as it rotates with a certain speed.
- The sides and the person have the same coefficients of static friction.

- We have to resolve the tensions into their x and y components.
- We have T 1 cos 45 and T 1 sin 45.
- The sum of all the forces must be zero.
- Since T 2 is completely horizontal, this means that T 1 cos 45 is T 2.
- The weight of T 1 must be balanced by the y component.

- The large mass is moving downward while the small mass is moving upward.
- The tension in the string is directed upward.

- We get ( M - m ) g /( M + m ) if we eliminate the tension T and solve for it.

- The downward force of gravity is opposed by the upward tension in the string.
- The upward net force must be achieved by doing a quick tip-to-tail graphical solution.

- The given values and solutions give us 35.7 N.

- The downward force of gravity is against this.
- The centripetal force F-c is produced by the combination.

- A sketch is shown.
- A downward net force must be achieved by doing a quick tip-to-tail graphical solution.
- 7,300 N upward is given by substituting values and solving for N.

- Net force is not acting on the puck when it is moving.
- The only net force that slows the puck down is the friction.

- The known numbers give a deceleration of -3.43 m/s 2.

- The distance the puck will travel before stopping is 58 m.

- The constant is 50 N/m.

- Hooke's law states that Fs is kx where x is the elongation in excess of the spring's natural length.
- The force is provided by the component of weight parallel to the incline.
- The answer is 0.646 m since this must be in excess of the spring's natural length.

- We drew a free-body diagram for the mass.

- The force that the 8 kilo mass exerts on the 2 kilo mass is represented by P. To find P, we use the known numbers.
- P is 16 N.

- The component of gravity is always given by the m g sin th.
- The force up the incline is at the same angle as the component parallel to it.

- The known numbers give a value of 3.17 m/s 2.

- The sum of all forces, in each direction, equals m a. T + 20 is 4 a, since the tension in the string will try to pull left.
- The tension pulls up against gravity.
- T - 18.6 is 2 a.
- We get T of 19.7 N by eliminating a.

- The known number gives us an acceleration of 0.67 m/s 2.
- On an incline, the normal force is given by mg cos th.
- The downward force of gravity against the incline was again opposed by this force.
- The two forces are combined in the downward direction.

The coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the coefficients of the

- N-1 is up and N-1 is down on the mass.

- The f-s is to the right on the 3- kilo mass and the left on the 7- kilo mass.

- The applied force, F, is acting on the 7- kilogram block.
- The force F must accelerate both blocks.

- Since we have horizontal motion, the normal force is equal to the combined weights, we can write.
- F is 84.3 N and we get it with Solving for F.

- The normal force on this surface is the weight of the block, as shown in the free-body diagram drawn in part a.

- From the vertical components, we can see that the N cos th is.

- We find that tan th is 1.59 and th is 57.8deg.

- The normal force is the same as the person and supplies the inward centripetal force.
- The person is slide down by Friction against the wall.
- To maintain equilibrium, the key is to be fast.

- The apparent weight of the mass at the equator is reduced by an upward force in the frame of reference.
- Only very sensitive scales can measure this effect.

- This is a very difficult question.
- The action and reaction pairs act on different objects.
- An object can move in a frame of reference if the applied force results in a net force.
- The force reaction can be experienced by a different object, which may or may not have its own acceleration.

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