Asked by Dani
Libby pulls a 3.9kg box to the right with a force of 200 n at a 45 degree angle relative to the horizontal along a frictionless surface. What is the acceleration of the box?
Answers
Answered by
Elena
F₀cosα=ma₀ (x)
a₀ (x) =F₀ cosα/m=200•0.707/3.9 =36.3 m/s²
F₀ sinα =ma₀ (y)
a₀ (y) = F₀ sinα/m=200•0.707/3.9 =36.3 m/s²
a(y)= a₀ (y)-g =36.3-9.8 =26.5 m/s²
a=sqrt(a₀ (x)² +a(y)²)=44.9 m/s²
a₀ (x) =F₀ cosα/m=200•0.707/3.9 =36.3 m/s²
F₀ sinα =ma₀ (y)
a₀ (y) = F₀ sinα/m=200•0.707/3.9 =36.3 m/s²
a(y)= a₀ (y)-g =36.3-9.8 =26.5 m/s²
a=sqrt(a₀ (x)² +a(y)²)=44.9 m/s²
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