A crate of mass 50kg is pulled up a rough inclined plane, inclined at an angle of 60° to the horizontal. Determine the total work done in moving the crate to the top of the inclined 200m high.

Wc = m*g = 50kg * 9.8N/kg = 490 N = Wt.

of crate.

Fc = 490N  60o = Force of crate.

Fp = 490*sin60 = 424.4 N. = Force

parallel to incline.

Fv = 490*cos60 = 245 N. = Force perpendicular to the incline.

Ff = u*Fv = u*245 = Force of friction.

Fap-Fp-Ff = m*a

Fap-424.4-u*245 = m*0 = 0

Fap = 424.4 + 245u. = Force applied.

L = 200m/sin60 = 231 m. = Length of

incline.

Work = Fap * L=(424.4+245u) * 231.

Explanation:

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

are their multiple choices ?

explanation: