Physic I

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tjbrewer | Elementary School Teacher | (Level 2) Associate Educator

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1) Work is energy transferred to an object by a force acting on the object, since energy is measured in Joules (J) work is also measured in Joules.  The formula for work, is `F xx d` .  The force on the elevator is 5500N, and its distance is 50.0m so the work done on the elevator is `5500 N xx 50.0 m=275,000`J

2) The hiker has a mass of 55.6 kg, and gravitational force on the surface of the earth is mass (in kg) x 9.8 `m/s^2` so Gravity has a force of `55.6 xx 9.8=` 544.88 N on the hiker.  If the hiker moves upward, Gravity does negative work, and that work is the force of gravity on the hiker, times his upward (negative) distance of 50 m, which is `544.88 xx -50=` -27,400J

3) Since work is energy transferred to an object by a constant force, when work ceases, the object should have kinetic energy equal to the work done on it (assuming no friction acts on the object).  The work done on the cart is the force applied (200N) times its distance (10m) `200 xx 10=2000` .  So the cart should have 2000J of kinetic energy. 

4) Work done on the car is energy given to the car.  The car was already in motion, so it already had energy.  The formula for an object's kinetic energy is `1/2mv^2` , filling in the car's mass (1208 kg), and velocity (10 m/s), we find that the car had `1/2(1208)(10^2)=1/2(1208)(100)=` 60,400J to begin with.  400,000J of work is done on the car, giving the car that much more kinetic energy, so the car now has `400,000+60,400=` 460,400J of Kinetic energy.  We solve the formula for v `K=1/2 mv^2-gt2K=mv^2-gt2K/m=v^2-gtv=sqrt(2K/m)` .  We fill in the values we have K=460,400. m=1208 and we find that the final velocity of the car is `v=sqrt(2(460400/1208))~~27.61` the final velocity is 27.61 m/s

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