work=force*distance
a) distance in four seconds:
d= 1/2 at^2=1/2 (9/18)4^2=4m
work=9*4 joules
b) the fourth second? figure work in from zero to three, and subtract from a) above.
A force of 9.0N is applied to a body at rest on a smooth horizontal surface.
If the body has a mass of 18.0kg and it moves in the direction of the force, find the work done in.
a. The first four second. b. The fourth second ? Urgent answers pls
12 answers
#bobpursely thank you very much for the reply
w = f * d
the mass is accelerated
... a = f / m = 9.0 / 18.0
d = 1/2 a t^2
... find the distances at the various times
the 4th sec is from 3 s to 4 s
the mass is accelerated
... a = f / m = 9.0 / 18.0
d = 1/2 a t^2
... find the distances at the various times
the 4th sec is from 3 s to 4 s
#scott .. Thanks for your reply
#bobpursley pls more enlightenment on the (b) aspect
#scott . you can also help me that thaty
b. You know the distance in 4 seconds. Figure the distance in 3 seconds. Subtract That is the distance in the fourth second.
Please how do I do that.. I am confused
How should I subtract it...
Please somebody should work it out for me
Please somebody should work it out for me
You know how to find the distance at 4 seconds
d = (1/2) a t^2 where t = 4
now do that again for t = 3
subtract to find the distance between the two points
d = (1/2) a t^2 where t = 4
now do that again for t = 3
subtract to find the distance between the two points
d= 1/2 at^2=1/2 (9/18)3^2 =2.25
Work =9*3
=27J ?
Am I correct
Work =9*3
=27J ?
Am I correct
9.0*2.25
= 20.25J ..
Am I right?
= 20.25J ..
Am I right?
2 answers