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いものとする。 問1 斜面に治って上昇する小物体 B の加速度の大きさ aa はどのように表されるか。正し いものを, 次の(1)〜(4)の中から一つ選...
Apr 25, 2024
いものとする。 問1 斜面に治って上昇する小物体 B の加速度の大きさ aa はどのように表されるか。正し いものを, 次の(1)〜(4)の中から一つ選びなさい。 (1) M−2msin⁡θ4m+Mg\frac{M-2 m \sin \theta}{4 m+M} g (2) 2(M−2msin⁡θ)m+4Mg\frac{2(M-2 m \sin \theta)}{m+4 M} g (3) M−2msin⁡θm+4Mg\frac{M-2 m \sin \theta}{m+4 M} g (4) 2(M−2msin⁡θ)4m+Mg\frac{2(M-2 m \sin \theta)}{4 m+M} g
Solution
a
Identify the forces acting on the small object B as it ascends the incline. These include the gravitational force mgmg acting downward, the normal force NN acting perpendicular to the incline, and the frictional force if any
b
Resolve the gravitational force into components parallel and perpendicular to the incline. The parallel component is mgsin⁡θmg \sin \theta and the perpendicular component is mgcos⁡θmg \cos \theta
c
Apply Newton's Second Law in the direction parallel to the incline. If there is no friction and assuming the mass of the object is mm, the net force acting on the object is (M−msin⁡θ)g(M - m \sin \theta)g, where MM is the mass causing the tension in the string
d
The net force is equal to the mass times the acceleration, so we have (M−msin⁡θ)g=(m+4M)a(M - m \sin \theta)g = (m + 4M)a
e
Solve for the acceleration aa to find the correct expression. a=(M−msin⁡θ)gm+4Ma = \frac{(M - m \sin \theta)g}{m + 4M}
C
Key Concept
Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Equation: F=maF = ma (Net force equals mass times acceleration)
Explanation
The correct expression for the acceleration of the small object B as it ascends the incline is derived by applying Newton's Second Law to the forces acting parallel to the incline. The net force is the difference between the tension and the component of the gravitational force parallel to the incline, and this net force is responsible for the acceleration of the object.
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