- #1
Any Help
- 79
- 2
1. Problem Statement:
A worker lifts a heavy crate vertically 1.0 m in 2.0 s. If she lifts the same crate the same distance in 1.0 s, the work was done by her is
delta x=a.t^2/2+v0.t
Fnet=ma Weight=mg
W=F.d
g=9.8N/kg
v0=0m/sec
delta x=a.t^2/2
t=2sec
1=a.4/2
a=0.5m/sec^2
F=m(0.5+9.8)=10.3m N
W=10.3m J
t=1sec
1=a1/2
a=2 m/sec^2
F=m(2+9.8)=11.8m N
W=11.8m J
then work change. How the correct answer stays the SAME I didn't get it.
A worker lifts a heavy crate vertically 1.0 m in 2.0 s. If she lifts the same crate the same distance in 1.0 s, the work was done by her is
A
the same.
B
two times as great.
C
half as great.
D
four times as great.
E
one-fourth as great.
F
zero.
Homework Equations
delta x=a.t^2/2+v0.t
Fnet=ma Weight=mg
W=F.d
g=9.8N/kg
The Attempt at a Solution
v0=0m/sec
delta x=a.t^2/2
t=2sec
1=a.4/2
a=0.5m/sec^2
F=m(0.5+9.8)=10.3m N
W=10.3m J
t=1sec
1=a1/2
a=2 m/sec^2
F=m(2+9.8)=11.8m N
W=11.8m J
then work change. How the correct answer stays the SAME I didn't get it.
Last edited by a moderator: