Asked by Me
A cable with 21.2 N of tension pulls straight up on a 1.10 kg block that is initially at rest. What is the block's speed (in m/s) after being lifted 2.42 m assuming that the local acceleration due to gravity is 9.80 m/s2?
Answers
Answered by
bobpursley
Net force= totalmass*(g+a)
solve for a.
then speed^2=2*acceleration*distance
solve for a.
then speed^2=2*acceleration*distance
Answered by
Hegde
We know that,
W = KE + PE ----> *
W = F x d
KE = 1/2 m v^2
PE = mgh
putting all this in *,
(F)(d) = 1/2 m v^2 + mgh
Rearranging,
v = sqrt[2[(F)(d)-(m)(g)(h)]/m]
Plugging in the values we get,
v = sqrt[2[(21.2N)(2.42m)-(1.1kg)(9.8m/s^2)(2.42m)]/1.1kg]
v = 6.77 m/s
W = KE + PE ----> *
W = F x d
KE = 1/2 m v^2
PE = mgh
putting all this in *,
(F)(d) = 1/2 m v^2 + mgh
Rearranging,
v = sqrt[2[(F)(d)-(m)(g)(h)]/m]
Plugging in the values we get,
v = sqrt[2[(21.2N)(2.42m)-(1.1kg)(9.8m/s^2)(2.42m)]/1.1kg]
v = 6.77 m/s
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