GRADE 11 PHYSICS

I have two Mechanical Energy questions that I am stuck on.

1) In the attachment below, the first page is the explanation we were given about kinetic energy. Then, there is the practice 4 question that has been highlighted. In the question, there is a hint given, in I don't understand. The second attachment underneath this is the answer to the practice question.

The thing I don't understand is where the 27J has come from. It is supposed to be the energy that will increase the speed of the vehicle, but when I attempt to solve for the same number (27J) using the Ek=1/2(mv^2) formula, I do not get that number. Even though the question is not asking for it, where is this number coming from?

The answers given here use the following sample data. Your data may be different, though your analysis should be the same. Sample mass = 75 kg Sample height of a ight of stairs = 3.5 In Sample time to walk the ight of stairs = 5.6 s Sample time to run the ight of stairs = 2.8 s (continued on next page) (a) Calculate your work done while walking and while running. The work done is the same in both cases: W=mgAh=(75 kg)(9.8m/s2)(3.5m)=2572.5.1 or 2600] Practice Question 4 (a) A toy car of mass 5.0 kg is travelling at a speed of 4.0 m/s. 27 J of work is done to increase the speed of the toy car. Calculate its final speed. Givens: m=5.0 kg v1 =4.0mls W =27 J Unknown: V , = ? 2 W Equation: W =-mv2 - m vi tv - 2 2 m 2 (27 J) + ( 40 m ) =5.2 m Substitute and Solve: V2=1 5 kg S SKinetic Energy When a force is applied to accelerate an object from speed in to speed V2, the work done on the object can be written as: W=F Ad =maAd Substituting a = 1122)] and A d = vavA t=%(v2 +vl )A t into this equation and simplifying yields: W = g m vi % m 113 Since this work is done to increase the speed of the object, the energy that is increased is the object's kinetic energy E. . . . . 1 The kinetic energy at any instant 18 therefore: E k =5 m V2 Note that the direction of the speed does not matter! (a) A toy car of mass 5.0 kg is travelling at a speed of 4.0 m/s. 27 J of workE- is done to increase the speed of the toy car. Calculate its nal speed. Hint: calculate the initial kinetic energy. Add the work done to nd the nal kinetic energy. Rearrange the equation for kinetic energy to nd the nal speed