Solution.
"F_p=50N;"
"\\varTheta=30^o;"
"F_f=40N;"
"d=3m;"
"W-?;"
"W=Fcos\\varTheta d;"
Traction force work: "W_p=50N\\sdot0.866\\sdot3m=130J;"
Work of friction force (angle is equal to 180o): "W_f=40N\\sdot(-1)\\sdot3m=-120J;"
Net work:
"W=W_p+W_f;"
"W=130J-120J=10J;"
Answer: "W=10J."
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