2) Assume the workman was able to get the sled moving. He is now pulls it 50[m] before stopping. How much work did the workman (through the use of the rope) do to the sled? A) 50,000sin(0) [J] XB) 50,000cos(0) [J] C) 50,000 [J] 200 [J] E) None of the above

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Chapter5: Newton's Law Of Motion
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2) Assume the workman was able to get the sled moving. He is now pulls it 50[m] before stopping.
How much work did the workman (through the use of the rope) do to the sled?
A) 50,000sin(0) [J]
XB) 50,000cos(0) [J]
C) 50,000 [J]
D) 200 [J]
E) None of the above
Transcribed Image Text:2) Assume the workman was able to get the sled moving. He is now pulls it 50[m] before stopping. How much work did the workman (through the use of the rope) do to the sled? A) 50,000sin(0) [J] XB) 50,000cos(0) [J] C) 50,000 [J] D) 200 [J] E) None of the above
The following information pertains to questions 1, 2, and 3. A workman pulls a sled
across an asphalt road using a rope at angle above the horizontal direction. The
workman pulls with 1,000[N] of force. The coefficients of friction of the sled with
the road are u=1.00 and u-0.70. The sled has mass m. The sled is initially at rest.
T7
1) Which expression gives the magnitude of fmax for the situation illustrated above?
A) mg
В)
mg – 1000cos(0)
C) mg+ 1000cos(0)
XD) mg – 1000sin(0)
E)
mg + 1000sin(0)
We know
max
= Hls N
N
Alo00 Sino
-> lo00 oso
N+ 1000 eino
N =
L000
sing
Put of N in eguetion(1)
Value
max
F ls (mg-loo0 Ling)
ls= 1-00 Cariven
(max
- lovo sine)
m (mg -
= 1:00
Ts
max
- 1000 sing
my
option o)
eptin
(D)
mg
loo0 sino is correct
Transcribed Image Text:The following information pertains to questions 1, 2, and 3. A workman pulls a sled across an asphalt road using a rope at angle above the horizontal direction. The workman pulls with 1,000[N] of force. The coefficients of friction of the sled with the road are u=1.00 and u-0.70. The sled has mass m. The sled is initially at rest. T7 1) Which expression gives the magnitude of fmax for the situation illustrated above? A) mg В) mg – 1000cos(0) C) mg+ 1000cos(0) XD) mg – 1000sin(0) E) mg + 1000sin(0) We know max = Hls N N Alo00 Sino -> lo00 oso N+ 1000 eino N = L000 sing Put of N in eguetion(1) Value max F ls (mg-loo0 Ling) ls= 1-00 Cariven (max - lovo sine) m (mg - = 1:00 Ts max - 1000 sing my option o) eptin (D) mg loo0 sino is correct
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