800 J pf work are done pushing a 40.0 kg crate 6.0 m along a horizontal surface with u = 0.30. If the crate started at rest, what speed does it reach
5 years ago
Answered By Bei Z
Step 1: From the work (800 J) and the distance (6.0 m) that the crate moved, the force applied on the crate can be found.
W=F*d 800=F*(6) so F=800/6=400/3 N
Step 2: From the F, we can find the acceleration a
F= ma 400/3=40*a so a=10/3 (m/s2)
step 3: Now it becomes a distance and velocity problem. The crate starts from a rest, which means that the initial velocity is 0; the distance is 6.0 m. From the equation correlated distance, acceleration and time, we can find the time for the moving.
d=v0t+(1/2)at2=(1/2)at2 (because v0=0)
6=(1/2)*(10/3)*t2
so t=(36/10)(1/2) sec
step 4: now time is known, accelerator is known, it is easy to get the final velocity.
5 years ago
Answered By Bei Z
Step 1: From the work (800 J) and the distance (6.0 m) that the crate moved, the force applied on the crate can be found.
W=F*d 800=F*(6) so F=800/6=400/3 N
Step 2: From the F, we can find the acceleration a
F= ma 400/3=40*a so a=10/3 (m/s2)
step 3: Now it becomes a distance and velocity problem. The crate starts from a rest, which means that the initial velocity is 0; the distance is 6.0 m. From the equation correlated distance, acceleration and time, we can find the time for the moving.
d=v0t+(1/2)at2=(1/2)at2 (because v0=0)
6=(1/2)*(10/3)*t2
so t=(36/10)(1/2) sec
step 4: now time is known, accelerator is known, it is easy to get the final velocity.
vf=v0+at=0+(10/3)*(36/10)(1/2)=2(10)(1/2)=6.32 m/sec
Hope this can help you!