1.
(a)Calculate the work done in lifting a 520 N barbell 2.4 m above the floor
(b)What is the potential energy of the barbell when it is lifted to this height?
2.
Suppose an automobile has 3100 J of kinetic energy.
(a)When it moves at twice the speed, what will be its kinetic energy?
(b)What's its kinetic energy at three times the speed?
2007-10-03 13:00:33 · 3 answers · asked by atlz1anonly 1 in Science & Mathematics ➔ Physics
w = mgh so (since F = ma and you already have that)
w = 520*2.4 = 1248 J and the potential energy is exactly the same.
Ek = (1/2)mv² so at twice the speed it will have
3100*4 = 12,400 J and at three times the speed it will have
3100*9 = 27,900 J.
Doug
2007-10-03 13:14:54 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
no calculator but ill try.
1.
(a) work = F * D
just multiply the two number you have together. the answer is in joules.
(b) i forgot the formula.
2.
(a) KE = .5MV^2
so its 4 times larger . (plug in 2 for V and you get 4 when you square it. then plu in 4 and you get 16.
(b) 9 times larger (again plug in 2 and you get 4 when you square it. plug in 6 and you get 36 when you square it.)
hope i helped.
2007-10-03 13:11:43 · answer #2 · answered by Anonymous · 0⤊ 0⤋
W = ?F?dx (F and dx are the two vectors, yet you probable do not care approximately that, right here we are conversing some million dimensions): W = ?Fdx --> W = ?(kx²)dx, from 0 to z, --> ?kz³ - 0 = ?kz³, that's C. you may divide with the help of three, because of the fact the mandatory of x² --> x³, so which you in basic terms divide with the help of the recent exponent (i.e. 3 thus)
2016-10-20 23:16:34 · answer #3 · answered by Anonymous · 0⤊ 0⤋