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A comprehensive set of exam questions and answers for phys 1114 chapter 6, covering topics such as simple machines, work, energy, and simple harmonic motion. It includes multiple-choice questions, true/false statements, and problem-solving exercises, offering a valuable resource for students preparing for exams.
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Simple Machine - CORRECT ANSWERS Any mechanical device that multiplies the effect of an applied force Lever - CORRECT ANSWERS By applying a small force at one end of a lever a larger force can be exerted on the rock at the opposite end Pulley - CORRECT ANSWERS Net result is that you can lift a weight a certain height by applying a force equal to only half the weight being lifted T/F When using a pulley you must pull the rope twice the distance the weight is lifted - CORRECT ANSWERS True What is the result for an ideal simple machine? - CORRECT ANSWERS Work output=work input Mechanical advantage - CORRECT ANSWERS Ratio of output force to input force Work - CORRECT ANSWERS Depends on both the strength of the applied force and the distance the object is moved Work formula - CORRECT ANSWERS W=Fd T/F Only the portion of the force that is perpendicular to the floor is used in computing work
Kinetic Energy Formula - CORRECT ANSWERS KE=1/2mv^ Negative Work - CORRECT ANSWERS the work done by a force acting in a direction opposite to the objects motion W=-Fd Potential Energy - CORRECT ANSWERS work is done but no KE is gained Potential Energy Formula - CORRECT ANSWERS PE=mgh T/F If force is applied to lift a crate, the gravitational potential energy of crate had decreased
a. What is the work done by this net force? b. What is the increase in kinetic energy of the mass? - CORRECT ANSWERS Net Force= 600 N Kinetic Energy=600 J *Work done= KE since there is no mention of frictional forces A spring with spring constant of 2 N/m is stretched a distance of 1 m from its original unstretched position. What is the increase of the potential energy of the spring? Give your answer in Joules. - CORRECT ANSWERS 1 J *PE=work done= Fd F=1/2kx-----this gives you F then plug it into PE=Fd Which takes more work: lifting a 2-kg rock to a height of 4 m without acceleration, or accelerating the same rock horizontally from rest to 10m/s? - CORRECT ANSWERS Accelerating the rock horizontally A 1-kg mass attached to a spring is pulled back horizontally across a table so that the potential energy of the system is increased from zero to 108 J. Ignoring friction, what is the kinetic energy of the system after the mass is released and has moved to a point where the potential energy has decreased to 77 J? Give your answer in Joules. - CORRECT ANSWERS 31 J subtract 108 and 77 A certain amount of work has been done in moving the mass of 1 Kg. from its initial position to a point where its potential energy (P.E.) is 108 J. At this point, all of the energy associated with the mass is P.E.; it is not moving, so its kinetic energy, K.E. is zero. Now when the mass is released, the P.E. is continuously converted to K.E: the P.E. decreases; the K.E. increases. ** The sum total of the energy remains constant. So, when the point is reached where the P.E. is down to 77 J, how much K.E. does the mass have now? The frequency of oscillation of a pendulum is 0.5 cycles/s What is the period of oscillation? Give your answer is seconds. - CORRECT ANSWERS 2 s
*P=1/f