a = F/m = 16.6/3.1
d = (1/2) a t^2
3.8 = (8.3/3.1) t^2
t = 2.01 seconds
v = a t = (16.6/3.1)(2.01)
A 3.1 kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 16.6 N. Find the speed after it has moved 3.8m.
4 answers
a = F/m = 16.6/3.1
d = (1/2) a t^2
3.8 = (8.3/3.1) t^2
t = 1.19 seconds
v = a t = (16.6/3.1)(1.19)
= 6.38 m/s
or do using change in Ke = work done
(1/2) 3.1 v^2 = 16.6*3.8
v = 6.38 m/s
d = (1/2) a t^2
3.8 = (8.3/3.1) t^2
t = 1.19 seconds
v = a t = (16.6/3.1)(1.19)
= 6.38 m/s
or do using change in Ke = work done
(1/2) 3.1 v^2 = 16.6*3.8
v = 6.38 m/s
i dont like physics for your information
you are given:
u: 0
s: 3.8
F: 16.6
m: 3.1
As F=ma, a=F/m, making a=16.6/3.1=5.35483871
kinematics equation:
v^2=u^2+2as
v^2=0^2+2(5.35483871)(3.8)
v^2=40.6967742
v=6.379402339 m/s
u: 0
s: 3.8
F: 16.6
m: 3.1
As F=ma, a=F/m, making a=16.6/3.1=5.35483871
kinematics equation:
v^2=u^2+2as
v^2=0^2+2(5.35483871)(3.8)
v^2=40.6967742
v=6.379402339 m/s