An object of mass 0.8kg moves due south on a horizontal frictionless plate plane with a speed of 0.5ms^-1. A horizontal force of magnitude 0.0060N directed due east acts on the object for 10s and is then removed. At the instant the force is removed what are the following?
The Velocity of the object?
The Displacement of the object relative to its position at the instant the force was applied (i.e. a distance and direction relative to where the object was at t = 0s)?
The work done on the object by the force?
2007-09-04 06:49:49 · 5 answers · asked by Alexander M 1 in Science & Mathematics ➔ Physics
The E/W change in momentum = F*t
= 0.06 kg m/s
u = 0.06 / 0.8 = 0.075 m/s east
Velocity = 0.5 m/s south, 0.075 m/s east
New KE of object = 0.4 * (0.5^2 + 0.075^2)
Old KE = 0.4 * (0.5^2)
Change in KE = 0.4 * (0.075^2) J = 0.00225 J
This is the work done on the object
2007-09-04 08:24:41 · answer #1 · answered by Dr D 7 · 0⤊ 0⤋
If force is due east then acceleration is also due east, so southward component of the velocity is unchanged while eastward component increase as a*t, where a=F/m=0.0075m/s^2 according to Newton's second law.
After 10 seconds eastward component of velocity will be 0.075m/s.
Total velocity can be found now from Pithagorean theorem
Sqrt(0.5^2 + 0.075^2)~0.5056m/s
Southward component of dispacement is 0.5*10=5m while eastward component is at^2/2=0.0075*100/2=0.375m. Total displacment according to Pithagorean theorem is Sqrt(5^2+0.375^2)~5.014m
The work done by force result to change in kinetic energy and equal to mv^2/2 - mv0^2/2 = 0.8(0.5^2 + 0.075^2)/2 - 0.8*0.5^2/2 = 0.8*0.075^2/2 = 2.25*10^-3 J.
2007-09-04 14:14:43 · answer #2 · answered by Alexey V 5 · 1⤊ 1⤋
vy = -0.5 m/s
from F = ma ==> a = F/m
acceleration(x direction) = 0.0060 N /0.8 ms^-2 = .0075 m/s^2
velocity in the x direction = a*t = 0.0075*10 = 0.075 m/s
velocity of the object = sqrt(vx^2 + vy^2) = sqrt(.075^2 + .5^2) = 0.51 m/s *****
displacement in the x direction
sx = 1/2*a*t^2 = 1/2*.075*10^2 = 3.75 m
displacement in the y direction
sy = vy * t = -.5*10 = -5 m
total displacement = sqrt(sx^2 + sy^2)
total displacement = 6.25 m****
angle = atan(sy/sx) = -53.1 degrees below the horizontal.****
since the force only acts in the xdirection
work done = F*sx = .0060 * 3.75 = 0.0225 J*****
2007-09-04 14:02:48 · answer #3 · answered by civil_av8r 7 · 0⤊ 1⤋
you have a problem with the force gained with a moving object
2007-09-05 11:58:42 · answer #4 · answered by marie c 1 · 0⤊ 0⤋
mv2=mv1+f*Dt so 0.8*v2 = 0.8* 0.5+ .0.006*10
v2= 0.46 /0.8=0.575m/s tan @== 0.06/0.4 so @
=0.1489rad
w=0.006*sin 0.1489 *0.575 *10=0.0051 joules
2007-09-04 14:10:58 · answer #5 · answered by santmann2002 7 · 0⤊ 2⤋