How much work is done on the block if a 10.0 kg block was accelerated at 5.0 m/s2 with a distance of 2.5 m across a frictionless table?
Given:
"m=10.0\\:\\rm kg"
"a=5.0\\:\\rm m\/s^2"
"d=2.5\\:\\rm m"
The force acting on a block
"F=ma=\\rm 10.0\\: kg*5.0\\: m\/s^2=50.0\\: N"The work done
"W=Fd=\\rm 50.0\\: N*2.5\\: m=125\\: J"
Comments
Leave a comment